# W.P.energy Final Colour

November 1, 2017 | Author: manan | Category: Force, Euclidean Vector, Power (Physics), Momentum, Heat

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Work, Power & Energy CONCEPT NO TES NOTES 01.

Concept of Energy

02 - 03

02.

Work: Definition and Calculation

03 - 12

03.

Work Energy Theorem and its Application

13 - 24

04.

Power

25 - 31

05.

Potential Energy and Energy Conservation

32 - 47

06.

More on Conservative forces

48 - 54

07.

A Note on Mass and Energy (Optimal)

55 - 59

08.

Miscellaneous Examples

60 - 87

WORK, POWER & ENERGY

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WORK, POWER & ENERGY

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WORK POWER & ENERGY Introduction: Any body (or an assembly of bodies) represents, in fact, a system of mass points, or particles. If a system changes with time, it is said that its state varies. The state of a system is defined by specifying the positions and velocities of all constituent particles. Experience shows that if the laws of forces acting on a system and the state of the system at a certain initial moment are known, the motion equations can help predict the subsequent behaviour of the system. However, an analysis of a system’s behaviour by the use of motion equations requires so much effort (due to the complexity of the system itself), that a comprehensive solution seems to be practically impossible. Moreover, such an approach is absolutely out of the question if the laws of acting forces are not known. Besides, there are some problems in which the accurate consideration of motion of individual particles is meaningless (example: gas). Under these circumstances the following question naturally comes up: are there any general principles following from Newton’s laws that would help avoid these difficulties by opening up some new approaches to the solution of the problem. It appears that such principles exist. They are called conservation laws. As we have already discussed, the state of a system varies in the course of time as that system moves. However, there are some quantities, state functions, which possess the very important and remarkable property of retaining their values with time provided you don’t change your frame. Among these constant quantities, energy, linear momentum and angular momentum play the most significant role. The laws of conservation of energy (There used to be a separate conservation law or more until Einstein unified it with energy), momentum and angular momentum fall into the category of the most fundamental principles of physics, (when you will study the law of conservation of linear momentum you will find that if is just a restatement ! of Newtons’ second law: F = d (mv )/dt applied to a system. Angular momentum too, is some thing similar, conservation of energy on the other hand, is a novel concept). These laws have become even more significant since it was discovered that they go beyond the scope of mechanics and represent universal laws of nature. In any case, no phenomena have been observed so far which do not obey these laws. They work reliably in all quarters: in the field of elementary particles, in outer space, in atomic physics and in solid state physics. Having made possible a new approach to treating various mechanical phenomena, the conservation laws turned into a powerful and efficient tool of research used by physicists. The importance of the conservation principles is due to several reasons: 1.

2.

The conservation laws do not depend on either the paths of particles or the nature of acting forces. Consequently, they allow us to draw some general and essential conclusions about the properties of various mechanical processes without resorting to their detailed analysis by means of motion equations. Since the conservation laws do not depend on the nature of the acting forces, they may be applied even when the forces are not known. In these cases the conservation laws are the only tools for research remaining. This is the present trend in the physics of elementary particles.

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3.

Even when the forces are known precisely, the conservation laws can help substantially to solve many problems of motion of particles. Although all these problems can be solved with the use of motion equations (and the conservation laws provide no additional information in this case), the utilization of the conservation laws very often allows the solution to be obtained in the most straight-forward and elegant fashion, obviating cumbersome and tedious calculations. We shall begin examining the conservation laws with the energy conservation law, having introduced the concept of energy and work.

Section - 1

CONCEPT OF ENERGY

Energy : It is possible to give a numerical rating, called energy, to the state of a physical system. The total energy is found by adding up contributions from characteristics of the system such as motion of objects in it, heat content of the objects (though that can also be attributed to mechanical vibrations of particles), and the relative positions of objects that interact via forces. The total energy of a closed system always remains constant. Energy can not be created or destroyed, but only transferred from one system to another. Energy comes in a variety of forms, and physicists didn’t discover all of them right away. They had to start somewhere, so they picked one form of energy to use as a standard for creating a numerical energy scale. One practical approach is to define an energy unit based on heating of water. The SI unit of energy is the joule, J, named after the British physicist James Joule. One joule is the amount of energy required in order to heat 0.24 g of water by 1°C. Note that heat, which is a form of energy, is completely different from temperature. In standard, formal terminology, there is another, finer distinction. The word heat is used only to indicate an amount of energy that is transferred, whereas thermal energy indicates an amount of energy contained in an object. Once a numerical scale of energy bas been established for some form of energy such as heat, it can easily be extended to other types of energy. For instance, the energy stored in one gallon of gasoline can be determined by putting some gasoline and some water in an insulated chamber, igniting the gas, and measuring the rise in the water’s temperature. (The fact that the apparatus is known as a bomb calorimeter will give you some idea of how dangerous these experiments are if you don’t take the right safety precautions). Here are some examples of other types of energy that can be measured using the same units of joules. Type of energy: • chemical energy released by burning • energy required to melt a solid substance food • raising a mass against the force of gravity

• •

energy required to break an object chemical energy released by digesting

nuclear energy released in fission, etc.

Textbooks often give the impression that a sophisticated physics concept was created by one person who had an inspiration one day, but in reality it is more in the nature of science to rough out an idea and then gradually refine it over many years. The idea of energy was tinkered with from the early 1800’s onwards, and new types of energy kept getting added to the list.

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To establish a new form of energy, a physicist has to 1. show that it could be converted to and from other forms of energy, and 2. show that it is related to some definite measurable property of the object, for example its temperature, motion, position relative to another object, or being in a solid or liquid state. For example, energy is released when a piece of iron is stored in water, so apparently there is some form of energy already stored in the iron. The release of this energy can also be related to a definite measurable property of the chunk of metal: it turns reddish-orange. There has been a chemical change in its physical state, which we call rusting. Although the list of types of energy kept getting longer and longer, it was clear that many of the types were just variations on a theme. There is an obvious similarity between the energy needed to melt ice and to melt butter, or between rusting of iron and many other chemical reactions. All the types of energy can be reduced to a very small number by simplifications.

Section - 2

WORK: DEFINITION AND CALCULATION

The concept of work: The mass contained in closed system is a conserved quantity, but if the system is not closed, we also have ways of measuring the amount of mass that goes in or out. We often have a system that is not closed, and would like to know how much energy comes in or out. Energy, however, is not a physical substance like water, so energy transfer can not be measured by the same kind of meter which we used to measure flow of water. How can we tell, for instance, how much useful energy a tractor can put out on one tank of gas? The law of conservation of energy guarantees that all the chemical energy in the gasoline will reappear in some form, but not necessarily in a form that is useful for doing farm work. Tractors, like cars, are extremely inefficient, and typically 90% of the energy they consume is converted directly into heat, which is carried away by the exhaust and the air flowing over the radiator. We wish to distinguish energy that comes out directly as heat from the energy that serves to accelerate a trailer or to plow a filed, so we defined a technical meaning of the ordinary work to express the distinction. Definition of work: Work is the amount of energy transferred into or out of a system, not taking into account energy transferred by heat conduction. [Based on this definition, is work a vector, or a scalar? What are its units?] The conduction of heat is to be distinguished from heating by friction. When a hot potato heats up your hands by conduction, the energy transfer occurs without any force, but when friction heats your car’s brake shoes, there is a force involved. The transfer of energy with and without a force are measured by completely different methods, so we wish to include heat transfer by frictional heating under the definition of work, but not heat transfer by conduction. The definition of work could thus be restated as the amount of energy transferred by forces. Work done by a constant force: Work done by a constant force is defined as product of the force and the component of the displacement along the direction of the force. WORK, POWER & ENERGY

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F m

S

4

F

m

fig. 5.1

Consider the situation shown in figure 5.1. A constant force F is applied on a block of mass m along the horizontal direction. If the block moves by a distance S on the horizontal surface on which it is placed, as shown in figure, then the work done by the force F is defined as ...(1). w = F ⋅S

F

θ

S

m

S

F

co s

θ

If the force and the displacement are not along the same direction, as shown in figure 5.2, then work done by force F is calculated by multiplying the force and the component of the displacement along the force, as shown in figure 5.3, therefore, for the given case, work done by force F is

F(const.) m

m

θ

S

fig. 5.2

m

fig. 5.3

w = F ⋅ S cos θ ⇒

w = F ⋅ S ⋅ cosθ

...(2)

Here you should note that work done by the force F can also be written as w = ( F cos θ ) ⋅ S

θ is the component of F along the displacement, as shown in figure 5.4. and we know that Fcosθ F(const.)

m

θ

S

F cos θ

m

work done by F = force along displacement × displacement fig. 5.4

Hence, work done by a constant force can also be defined as the product of the displacement and the component of the force along the displacement. In vector form equations (1) and (2) can be generalized as ! ! w = F ⋅S

...(3)

Hence, work done by a constant force is the scalar (dot) product of the force and the displacement. Here WORK, POWER & ENERGY

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! ! I would like to emphasize that S is the displacement of the point of application of the force F . ! Note: For a constant force F , in equation (3) you should notice the following: ! ! • When F ⊥ S, i.e., θ = 90°, w f = 0. •

When θ ∈ (0,90°), w f > 0.

When θ ∈ (90°,180°), w f < 0.

When θ = 180°, w f = −F ⋅ S

When θ = 0°, w f = F ⋅ S

! ! In a closed path work done by a constant force is zero. (∵ S = 0). ! ( w f denotes work done by the constant force F and S is the magnitude of the displacement S ) ! Equation (3) refers only to the work done on the particle by a particular force F . The work done on the particle by the other forces must be calculated separately. The total work done on the particle is the sum of the work done by the separate forces. ! When θ is zero, as shown in figure 5.5, the work done by F is F m simply F × S, in agreement with equation (1). Thus, when a F S (θ = 0°) S constant horizontal force draws a body horizontally, or when a constant vertical force lifts a body vertically, the work done by fig. 5.5 the force is the product of the magnitude of the force by the distance moved.

When θ is 90°, as shown in figure 5.6, the force has no component in the direction of motion. That force then does no work on the body. For instance, the vertical force holding a body a fixed distance off the ground does no work on the body, even if the body is moved horizontally over the ground. Also the centripetal force acting on a body in motion does no work on that body because the force is always at right angles to the direction in which the body is moving.

F m

F

S

(θ = 90°)

S fig. 5.6

Of course, a force does no work on a body that does not move, for its displacement is then zero. As I have already mentioned, the work done by a constant force can be calculated in two different ways: Either we multiply the magnitude of the displacement by the component of the force in the direction of the displacement or we multiply the magnitude of the force by the component of the displacement in the direction of the force. These two methods always give the same result. Work is a scalar, although the two quantities involved in its definition, force and displacement, are vectors. We define the scalar product of two vectors as the scalar quantity that we find when we multiply the magnitude of one vector by the component of a second vector along the direction of the first. Equation (3) shows that work is such a quantity.

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Work can be either positive or negative. If the particle on which a force acts has a component of motion opposite to the direction of the force, the work done by that force (Fig. 5.7) is negative. This corresponds to an obtuse angle, between the force and displacement vectors. For example, when a person lowers an object to the floor, the work done on the object by the upward force of his hand holding the object is negative.

(θ > 90°)

F θ m

S fig. 5.7

Our special definition of the word work does not correspond to the daily usage of the term. This may be confusing. A person holding a heavy weight at rest in the air may say that he is doing hard work- and he may work hard in the physiological sense-but from the point of view of physics we say that he is not doing any work. We say this because the applied force causes no displacement. The word work is used only in the strict sense of equation (3). The unit of work is the work done by a unit force in moving a body a unit distance in direction of the force. In the mks system the unit of work is 1 newton-meter, called 1 joule. Work done by gravity (near earth surface): Near the earth’s surface the gravitational force acting on a body due to attraction of the earth can be assumed to be constant. Therefore work done by gravity can be calculated using equation (3), whatever be the path of motion of the body. 1

1

!

!

θ

mg

mg

!

mg

path

!

h

path

mg h

S

!

S cosθ

mg

S

S cosθ

2 fig. 5.8

fig. 5.9

2

Consider the situation shown in figure 5.8. A particle of mass m is moved on an arbitrary path in a vertical plane. As the particle is moved from point 1 to point 2, its weight acting on it at different positions during its motion is also shown in figure. From figure it is clear that work done by gravity can be found by multiplying mg (magnitude of the force) by the downward displacement h of the body because the component of the displacement of the body along the force is h only. If you are not satisfied with the discussion above and want to calculate it mathematically, then you can proceed according to the following way: Work done by gravity when particle moves from point 1to point 2 as shown in figure 5.9 is ! ! w g = mg ⋅ s ! = mg ⋅ s cos θ [∵ mg is constant] = mg. h

= weight of the particle × downward displacement of the particle

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Hence, we have got the same result as we had predicted earlier. Therefore, when a particle comes down by a distance h through any path, work done by gravity is given as w g = mgh (downward motion)

...(4)

You should note that this work done is independent of the path of the particle. You should also note that “mgh” is work done by gravity only, not by all forces acting on the particle when it moved from point 1 to point 2. What would be the work done by gravity when the particle moves from point 2 to point 1? In this case the displacement of the particle has a component of length h in the vertical direction but this component is in the ! opposite direction of mg , hence, work done by gravity for this case is w g = −mgh (upward motion)

...(5)

You should note that this expression is also independent of the path followed by the particle while moving from point 2 to point 1. Now, let me propose a common equation for the equations (4) and (5). When a particle of mass m moves near earth surface, work done on it by gravity is given by w g = −mg∆h

...(6)

where ∆h is the change in height of the particle. If the particle goes up by a height h then ∆h = +h and wg = −mgh , which is in accordance with equation (5). If the particle comes down by a height h then ∆h = –h and work done by gravity, wg = + mgh, which is in accordance with the equation (4). Hence, for all cases we can use equation (6). The only restriction while using equation (4), (5) or (6) is that ‘mg’ must be uniform over the path for which work done is required. Note that if the particle moves from 1 to 2 and then from 2 to 1, then work done by gravity is zero. Hence, we can say that work done by gravity in a closed path is zero. Work done by a variable force: ! θ ds When the force varies over the path of motion of a particle (it 2 • can vary in magnitude or direction or both), then to calculate work done by the force we divide the path of the particle in many infinitesimally small intervals and calculate the work done in each interval and by adding the work done for all these small intervals we get the work done over the segment of the path 1 ! under consideration. In figure 5.10, such an interval, d s , is fig. 5.10 shown. As the interval is very small, we can assume that ! in this interval force is constant and hence work done by F in this interval can be written as ! ! ...(7) dw = F ⋅ ds ! where ds is the displacement of the particle for this! interval. As the particle moves from point 1 to point 2 on the path shown above, the net work done by the force F is calculated by adding work done in all subintervals of the ! path from point 1 to point 2. Therefore, net work done by F is WORK, POWER & ENERGY

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w = sum of all dw's

∫ dw

w=

! ! w = ∫ F ⋅d s 2

...(8)

1

You should note that when you apply equation (8) for a constant force you get the same result as given by equation (3). ! ! For a one dimensional case equation (8) can be modified by replacing d s with dx and replacing F with F(x). Therefore, we get, xf

w=

F ( x) ⋅ dx

...(9)

xi

For a 3-dimensional case equation (8) can be expanded as follows: fin

w=

! ! F ⋅ ds

in fin

∫ ( Fxiˆ + Fy ˆj + Fz kˆ ).(dxiˆ + dyjˆ + dzkˆ )

=

in

fin

∫ ( Fx ⋅ dx + Fy ⋅ dy + Fz ⋅ dz )

=

in xf

w=

yf

Fx ⋅ dx +

xi

zf

F y ⋅ dy +

yi

∫ Fz ⋅ dz

zi

...(10)

! If components of F along x, y and z are constant, then equation (10) reduces to

w = Fx ⋅ ∆x + F y ⋅ ∆y + Fz ⋅ ∆z

...(11)

There is an alternate way of looking at work done by variable forces which would prove to be a powerful method ! as we will proceed with the topic. Suppose we consider F as the sum of its components along the tangent and the normal to the path of motion of the particle. In that case equation (8) gives fin

w=

!

!

∫ F ⋅d s

in

fin

=

in WORK, POWER & ENERGY

( Ft ⋅ tˆ + Fn ⋅ nˆ ) ⋅ d s!

tˆ and nˆ are unit vectors along the tangent   and the normal to the path, respactively,     at the point under consideration  www.locuseducation.org

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=

!

9

!

∫  Ft (tˆ ⋅ d s ) + Fn (nˆ ⋅d s ) 

in

fin

w=

! ! ∵ nˆ ⊥ # d s and tˆ ⋅ d s = ds   

Ft ⋅ ds

in

...(12)

Hence, work is done only by the tangential component of a force.

Example – 1

WORK DONE BY SPRING FORCE

A light spring of spring constant k is stretched from an initial deformation xi to a final deformation x f quasistatically (i.e., equilibrium is always maintained). Find the work done by (a) spring force

(b) external agent

Solution: In figure 5.11 the spring is shown at some arbitrary deformation x. As the spring always exerts restoring force, at the shown moment it is applying a force in the opposite direction of x. Therefore, Fsp = −kx (always true) where k is the spring constant of the spring.

natural length

xi xf x dx

Fext

Fsp fig. 5.11

To increase the deformation the external must exert a force in the direction of x and the force exerted by the external agent, Fext , must have a magnitude equal to that of Fsp because equilibrium is always maintained. Therefore,  true only if equilibrium  Fext = + kx    is maintained  If the spring is further deformed by dx from the shown position then work done by the spring force is dwsp = (−kx) ⋅ dx

and work done by external agent is dwext = (+kx) ⋅ dx While deforming the spring from xi to x f , we get, total work done by the spring, xf

wsp = ∫ dwsp = −k ∫ x ⋅ dx xi

⇒ WORK, POWER & ENERGY

1 1  wsp = −  kx 2f − kxi2  2 2 

...(13) www.locuseducation.org

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and total work done by external agent xf

wext = ∫ dwext = +k ∫ x ⋅ dx xi

wext =

1 2 1 2 kx f − kxi 2 2

...(14)

Initially if the spring has its natural length then substituting xi = 0, equation (13) and (14) give

and

1 w sp = − kx 2f 2

...(15)

1 wext = + kx 2f 2

...(16)

Note: You must notice the following •

Equations (13) and (15) are true for all cases but equations (14) and (16) are true only if equilibrium is maintained.

Work done by the spring is independent of path, it depends only upon initial and final position. [See equation (13) and (15)]

• •

You should note that ∫ F ( x) ⋅ dx denotes area under F(x)

F(x)

curve when it is plotted against x. From figure. 5.12,

xi

– kxi

= A1 + A2

– kxf

xf A2

1 = (−kxi )( x f − xi ) + (−kx f + kxi )( x f − xi ) 2 1 1  = −  kx 2f − kxi2  2 2 

x

A1

fig. 5.12

Wsp

F(x)= – kx =Fsp

If we put xi = x f in equation (13), then we get wsp = 0. Therefore, we can say that work done by spring force in a closed path is zero.

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Work Done by Two or More than two forces: ! ! ! Suppose there are n forces F1, F2,........., Fn acting on a particle of mass m, as shown in figure 5.13. If the particle ! suffers a displacement ds in the next time interval dt, then the net work done on the particle is the sum of work done by all the forces acting on it. Hence, we have net work done on the particle ! dw = dw1 + dw2 + dw + ....... + dwn ! F3 F2 ! ! ! ! ! ! ! ! = F1 ⋅ ds + F2 ⋅ ds + F3 ⋅ ds + ....... + Fn ⋅ ds ! ! ! F1 ! m where dwi = Fi ⋅ ds is work done by the force Fi . For a finite displacement, net work done on the particle is w = w1 + w2 + ...... + wn

!

...(17)

! ! ! ! ! ! = ∫ F1 ⋅ ds + ∫ F2 ⋅ ds + .......... + ∫ Fn ⋅ ds

Fn fig. 5.13

! ! ! ! = ∫ F1 + F2 + ...... + Fn ⋅ ds

(

! ! w = Fnet ⋅ ds

)

...(18)

Therefore, if many forces are acting on a particle, then we have two ways to find out net work done on the particle: 1.

find work done by each force individually and then find the sum of work done by all forces;

2.

find the net force, sum of all forces, acting on the particle and then find work done by this net force.

When you are applying the above approach for a particle then no attention is needed but while applying this concept on a system of particles or on a body you need to be cautious. If points of application of different forces have different displacements, then equation (17) can not be reduced to equation (18). Hence, in such a case you are left with only one option and that is of using equation (17).

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TRY YOURSELF - I Q. 1

No work is done by a force on an object if (a) the force is always perpendicular to its velocity (b) the force is always perpendicular to its acceleration (c) the object is stationary but the point of application of the force moves on the object (d) the object moves in such a way that the point of application of the force remains fixed.

Q. 2

Find the work a boy of weight 55 kg has to do against gravity when climbing from the bottom to the top of a 3.0 m high tree. (g = 10 m/s²).

Q. 3

A body is thrown on a rough surface such that friction force acting on it varies linearly with distance traveled as f = ax + b. Find the work done by the friction on the box if before coming to rest the box travels a distance s.

Q. 4

A force is given by F = kx 2 , where x is in meters and k = 10 Nt/m2 . What is the work done by this force when it acts from x = 0 to x = 0.1 m?

Q. 5

A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to : (a) s (c) s

Q. 6

Q. 7

(b) s² (d) ln s

A force F acting on a particle varies with the position x as shown in figure. Find the work done by this force in displacing the particle from (a) x = −2m to x = 0 (b) x = 0 to x = 2m.

F(n) 10 –2 2

x(m)

–10

Two unequal masses of 1 kg and 2 kg are attached to the two ends of a light inextensible string passing over a smooth pulley as shown in figure. If the system is released from rest, find the work done by the string on both the blocks in 1 s. (Take g = 10 m/s²)

1 kg 2 kg

Q. 8

Q. 9

! ! A particle moves from a point r1 = (2m)iˆ + (3m) ˆj to another point r2 = (3m)iˆ + (2m) ˆj during which a \$! certain force F = (5 N )iˆ + (5 N ) ˆj acts on it. Find the work done by the force on the particle during the displacement. A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring.

WORK, POWER & ENERGY

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Section - 3

13

WORK ENERGY THEOREM AND ITS APPLICATION

As we have already discussed that the normal component of a force does no work, the work is done only by the tangential component of a force. The tangential component is related to rate of change of speed of the particle. By dv the Newton’s 2nd law, ∑ Ft = m , where v is speed of the particle and ∑ Ft denotes the sum of tangential dt components of all forces acting on the particle. We can now think of speed as a function of the distance s measured along the curve (as shown in figure 5.14) and apply the chain rule for derivatives: v = ds dt

dv dv ds dv = ⋅ = v⋅ dt ds dt ds

∑ Ft = m ⋅ v ⋅

s

dv ds starting point

where we have used the fact that fin

wnet =

=

or

wnet =

fig. 5.14

ds is just the speed v. The work done by resultant force is thus dt

∑ Ft ⋅ ds =

fin

dv

∫ mv ⋅ ds ⋅ ds

in

in

fin

fin

in

in

∫ mv ⋅ dv = m ∫ v ⋅ dv 1 1 mv 2f − mvi2 2 2

...(19)

1 mv 2 has the same unit as that of work and hence it is a form of energy. As it depends upon the 2 speed of the particle, it is defined as the kinetic energy, k, of the particle. Hence equation (19) can be rewritten as

The quantity

wnet = k f − ki

wnet = ∆k

...(20)

Therefore, net work done on a particle is equal to the change in kinetic energy of the particle. This theorem is known as the work energy theorem. If positive work is done on a particle then kinetic energy of the particle increases. If work done on a particle is negative then kinetic energy of the particle decreases. If no work is done on a particle then kinetic energy of the particle remains unchanged and hence speed of the particle remains constant. At this juncture you should recall the WORK, POWER & ENERGY

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very basic definition of work. We had defined it as energy transferred by forces and here the work energy theorem proves this definition. Now, it is clear that all the work done on a body goes in increasing the kinetic energy of the body. Using equation (12) and (20), we can get

dw = dk = Ft ⋅ ds dk ds

Ft =

...(21)

Therefore, the derivative of kinetic energy with respect to distance covered along the path gives the tangential component of the net force acting on a particle. The normal component of the net force can be found by multiplying the mass of the particle by the centripetal acceleration of the particle. We have already learnt that, in order to change the speed, we need a tangential component of the force (in circular motion). It is in accordance with equation (21). If the tangential force is zero, then no work is done on a particle and hence its kinetic energy remains unchanged or we can say that the speed of the particle remains unchanged. Therefore to change the speed there must be a component of the net force along the tangent to the path and this is exactly what we discussed while covering non uniform circular motion.

Example – 2 A small ball released from rest from a height h on a smooth surface of varying inclination, as shown in figure 5.15. Find the speed of the ball when it reaches the horizontal part of the surface. Solution: Let the speed of the ball when it reaches the horizontal part of the surface be v0 , as shown in figure 5.16(b).

h

smooth fig. 5.15

N m v0

v mg

fig. 5.16(a)

fig. 5.16(b)

The position of the ball at some arbitrary point of the inclined part of the surface is shown in figure 5.16(a). During the entire journey there are only two forces acting on the ball: (1) Its weight; (2) Normal contact force from the surface. Applying the work energy theorem between the moments when it was released from the rest and when it reaches the horizontal surface, we get. WORK, POWER & ENERGY

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wnet = ∆k ⇒

wg + w N

mgh + 0 =

0

= kf − k i

15

 wg → work done by gravity;     wN → work done by normal contact force 

0

from equation (6),  w = mgh   g 

1 2 mv0 − 0 2

v0 = 2 gh

...(22)

As the normal contact force is always perpendicular to the direction of motion, work done by it is zero. When a ball falls freely from rest by a distance h, its speed is the same as obtained in equation (22). During free fall the only force doing work is gravity.

Example – 3 m

In the previous problem suppose that an uncompressed spring is fixed on the horizontal part of the surface, as shown in figure 5.17. Now, if the particle is released from rest from the position shown in the figure, find the maximum compression in the spring if its spring constant is k.

h k

fig. 5.17

Solution: When compression in the spring is maximum, speed of the ball must be zero at that moment, as shown in figure 5.18 (c). If x0 be the maximum compression in the spring then applying work energy theorem on the ball between the instants when it was released from rest and when the compression in the spring is maximum, we get N N

m

x0

v

v0 v=0

mg

fig. 5.18(a)

fig. 5.18(b)

fig. 5.18(c) mg

wnet = ∆k ⇒

wg + w + wsp = k f N

1 +mgh − kx02 = 0 2

x0 =

0

0

− k i

0

wsp is work done by spring 

2mgh k

You should try to prove that the ball would bounce back to the same point from which it was released. WORK, POWER & ENERGY

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Example – 4 In the previous example suppose the spring is removed from the horizontal part of the surface and this part is made rough. If the coefficient friction between the rough part of the surface and the ball is µ, find the distance covered by the ball on the horizontal part of the surface before it comes to rest. Solution: Let the ball comes to rest having covered a distance s on the horizontal part of the surface. During the motion of the ball only three forces act on the ball: (1) Gravity; (2) Normal contact force; (3) Frictional force. Applying work energy theorem between the instants when the ball was released from rest and the ball comes to rest on the horizontal part, we get,

N

v mg

Fig. 5.19 (a)

wnet = ∆k ⇒

wg + wN + w fr = k f − ki

+mgh + 0 − µ mg ⋅ s = 0 − 0

s=

h µ

N sm oo th

µmg

v rough

Fig. 5.19 (b)

mg

Example – 5 A block of mass m suspended vertically by a light spring of spring constant k is released from rest when the spring is in its natural length state, as shown in figure 5.20. Find the maximum elongation in the spring.

k

natural length

Solution: As the block is released from rest it falls downwards due to its own weight but as it moves downwards its m downward acceleration decreases due to the upward “released from rest” force exerted by the spring. Initially the spring force Fig. 5.20 is zero, hence the downward acceleration is maximum (=g). But as the spring elongates the net downward force decreases and hence the downward acceleration also decreases. After some time the upward spring force becomes equal to the weight of the block. This position is known as equilibrium position because net force in this position is zero. You should note that at this position speed of the block is not zero. In fact, at this position speed of the block is maximum, because before arriving to this position spring force is smaller than the weight of the block and hence the speed of the block is increasing in the downward direction. WORK, POWER & ENERGY

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When the block reaches the equilibrium position, the net force on it becomes zero and hence its acceleration also becomes zero. Therefore, speed becomes maximum because it can not increase further. Although acceleration of the block is zero in this position, it will continue moving in the downward direction due to its downward velocity.

natural length position

v=0 kx a=g

mg a

x x0

v(increasing kx0

mg k

2mg 2x0= k =xmax

x

speed)

a=0

a

mg

kx

vmax

equilibrium position v (decreasing speed)

mg fig. 5.21

2kx0=2mg a=g v=0 mg

maximum elongation position

As the block passes the equilibrium position, the spring force exceeds the weight of the block and hence the block starts decelerating and then after moving some distance with decreasing speed, it momentarily comes to rest. Obviously when the ball comes to rest, elongation in the spring is maximum. If xmax is the maximum elongation in the spring then applying work energy theorem on the block between the moment when it was released and when the elongation is maximum, we get wnet = ∆k ⇒

wg + wsp = k f − ki

1 2 +mg ⋅ xmax − kxmax = 0−0 = 0 2

2mg k

xmax =

xmax = 2 x0

where x0 is the elongation in the spring in the equilibrium position. Note: • •

At x = xmax block momentarily comes to rest. At this position velocity of the block is zero but acceleration is not zero. It has an upward acceleration of magnitude g, as shown in figure 5.21.

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When elongation is maximum in the spring its speed is zero which can be proved very easily if you proceed mathematically rather than analyzing exactly what’s happening there. When elongation in the spring is maximum, distance of block from natural length state, x, as shown in figure 5.21, is also maximum. Hence, we have dx =0 dt

v = 0.

Example – 6 A small ball of mass m is suspended by means of a light thread of length l. When the ball is hanging vertically it is given a horizontal speed u, as shown in figure 5.22. Find the speed of the ball and tension in the thread supporting the ball when the thread makes an angle θ with the vertical.

O

Solution: At some angular displacement θ, the situation is shown in figure 5.23. Applying work energy theorem on the ball, we get, wall = ∆k ⇒

wT + wg = k f − ki

0 + wg =

u

m fig. 5.22 O

ω θ

1 2 1 mv − mu 2 2 2 1 2 1 mv − mu 2 2 2

−mg ∆h =

−mg.x =

v 2 = u 2 − 2gx

v 2 = u 2 − 2 gl(1 − cos θ )

T ⊥ # V  

v l

2

T

V

l x

1 2 1 mv − mu 2 2 2

θ

s

mg x = l – l cos θ = l (1− cos θ )

(i)

fig. 5.23

v = u 2 − 2 gl(1 − cos θ )

At the position shown in figure 5.23, the ball is moving in a vertical circle of radius l and has a speed v, therefore, its radial acceleration is v 2 /l. Applying Newton’s 2nd law along the radial direction at this moment, we get Fnet = ma ⇒

T − mg cos θ = m

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*

19

mu 2 − 2mg (1 − cos θ ) l

T = mg cos θ +

u2 T = 3mg cos θ − 2mg + m l

[using equation (i)]

You should notice that we had done this problem in the previous chapter also (CIRCULAR MOTION). That time we solved this problem by analyzing the varying tangential acceleration of the ball.

Example – 7 m h ot

sm o

A small ball is released from the top of a smooth hemispherical surface fixed on a horizontal plane as shown in figure 5.24. If m be the mass of the ball and R be the radius of the hemispherical surface, find the speed of the ball and normal contact force between the ball and the hemispherical surface as a function of θ, where θ is the angle between radial position of the ball with respect to the centre of the hemisphere and the vertical direction.

R

fig. 5.24

o

Solution: The position of the ball when it moves through an angle θ is shown in figure 5.25. As the ball is moving on a circular path of radius R, at the shown position it has a centripetal acceleration of magnitude v 2/R. Although, I am not discussing about the tangential acceleration of the ball at the shown moment, you should not forget that it is also present. s N Applying Work Energy Theorem, we get, x h wN + wg = k f − kki

⇒ ⇒

m

sm

wall = ∆k

ot

0 − mg ⋅ ∆h = −mg (− x) =

θ

1 2 mv − 0 2

2

v

R

θ mg

v

fig. 5.25

1 2 mv 2

v 2 = 2 gx

v = 2 gx

...(i) [ x = l (1 − cos θ )]

Applying Newton’s 2nd law along the radial direction, we get. Fnet = ma ⇒

mg cos θ − N =

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mv 2 = 2mg (1 − cos θ ) l

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N = 3mg cos θ − 2mg

⇒ *

20

Notice that at θ = cos −1(2/3), N becomes zero. What is the physical significance of this θ ? What would happen when the ball passes this position?

Example – 8 The kinetic energy of a particle moving along a circle of radius R depends upon the distance covered as k = α s 2 , where α is a constant. Find the magnitude of the force acting on the particle as a function of s. Solution: Let us first find the tangential and normal components of the net force acting upon the particle, then we can find the net force by adding these two components. If Ft be the tangential component of the net force, then Ft =

dk d (α s 2 ) = ds ds

= 2α s

If Fn be the normal component of the net force, then Fn = m

v2 R

 2  1  =    mv 2   R  2 

=

2 (α s 2 ) R

=

2α s 2 R

[∵ k = α s 2 ]

Fnet = Ft 2 + Fn2  2α s 2  = ( 2α s ) +    R 

2

2

s = 2α s 1 +   R =

WORK, POWER & ENERGY

2

2α s R2 + s2 R

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Example – 9 A particle of mass m starts moving so that its speed varies according to the law v = α s , where α is a positive constant, and s is the distance covered. Find the total work performed by all the forces which are acting on the particle during the first t seconds after the beginning of motion. Solution: Using work energy theorem, we have, work done by all forces wall = ∆k (change in K.E.) = k f − ki =

1 2 mv − 0 2

=

1 2 mv 2

=

1 mα 2 s 2

[∵ initially s = 0 ⇒ v = 0 ⇒ k = 0]

...(i)

So, now we have to find s as a function of t. We have ds =v dt

ds =α s dt

ds = α ⋅ dt s s

∫ 0

t

ds = α ∫ dt s 0 s

2 s =α t 0

2 s = αt

s=

α 2t 2 4

...(ii)

mα 2s wall = 2

[Using (i)]

mα 4t 2 8

[Using (ii)]

=

WORK, POWER & ENERGY

t

0

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Example – 10 A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as an = α t 2, where α is a positive constant. Find the time dependence of the work done by all the forces. Solution: We have,

an = α t 2 ⇒

v2 = αt2 R

v 2 = α Rt 2

Work done by all forces,

...(i)

wall = ∆k = k f − ki

=

1 2 mv − 0 2

=

1 2 mv 2

wall =

mα Rt 2 2

[Using equation (i)]

Example – 11 A chain of mass m and length l rests on a rough surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the overhanging part equals 1/3 of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table? Solution: When the chain has fallen by a distance x its position on the table is shown in figure 5.26. At this instant friction is opposing the motion of the chain and is acting towards the left. If µ be the friction coefficient between the chain and the table and m′ be the mass of the part of the chain on the surface of the table, then friction force is 2l/3 f = µm ' g x  m  2l  = µ   − x g  l 3

l/3 fig 5.26

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µ mg  2l   − x l 3 

=–

In the next infinitesimally small time interval dt if the chain slides further by dx, then work done by the frictional force can be given as dw = − f ⋅ dx

=

µ mg  2l   − x  ⋅ dx l 3 

Therefore, till the moment when the chain leaves the table surface completely (i.e., value of x becomes 2l/3), net work done by friction forces is w = ∫ dw

=−

µ mg l

x = 2l/3

∫ 0

 2l   − x  ⋅ dx 3 

2 µ mg  2l 2l ( 2l/3)  =−  × −  l  3 3 2 

=−

µ mg (2l/3)2 l 2

2 = − µ mgl 9

You should note that µ was not given in the question. You have to find it on your own. (Of course you can find it from the statement of the question). [Suppose you have to find out the speed of the chain at the moment it just leaves the table, then that can be found by using the result above in the work energy theorem. In this case there are only three forces acting on the chain: gravity, friction and normal contact force from the table. Normal contact force is not doing work so you need work done by friction and gravity only to find the change in K.E. of the chain, which would lead you to the final speed of the chain. And hence in this way you can avoid complicated equations and their solutions which you’d have encountered if you would have chosen methods learnt in chapter NEWTON’S LAWS OF MOTION.]

Example – 12 In the previous example find the work done by gravity (on the chain) for the same duration. Solution: For the same time interval dt which we considered in the previous example, work done by gravity on the chain is ml  dwg =   + x   g ⋅ dx   l 3 = WORK, POWER & ENERGY

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Hence, net work done on the chain by gravity till the moment it leaves the table completely is wg = ∫ dwg

=

=

mg l mg l

x = 2l/3

∫ 0

l   + x  ⋅ dx 3 

 l 2l (2l/3) 2   ⋅ +  2  3 3

mg  2l 2 2l 2  = +   9  l  9 =

Note:

2l/3 x l/3

fig 5.27

x dx

4 mgl 9

In the next topic we will study the concept of CENTRE OF MASS. When you are familiar with that concept, you find the work done by gravity in a much easier way, although the method discussed here is also simple and easy.

If v be the speed of the chain when it just leaves the horizontal surface, then applying work energy theorem on the chain between the moments when it started sliding over the horizontal surface and when it just left the surface, we get wall = ∆k ⇒

wg + w f + wN = k f − ki

1 2 4   2   mgl  +  − µ mgl  + (0) = mv − 0 2 9   9 

4 2 1 mgl − µ mgl = mv 2 9 9 2

Put the value of µ and then solve the above equation to get v.

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Section - 4

25

POWER

Power: The rate of doing work by a force with respect to time is defined as power developed by that force. We know that when a force does work on a body it transfers energy to that body, therefore, power developed by a force is the rate of energy transfer by that force. Therefore, for power, P, we can write P=

dw dt

...(23)

! ds! ! ! P =F⋅ = F ⋅v dt ! ! P = F ⋅v

⇒ ⇒

[∵

! ! dw = F ⋅ d s ]

...(24)

Therefore, power developed by a force is the scalar product of the force and the velocity of point of application of the force. The unit of power is Joule/sec which is defined as “watt”. It is obvious that when force is perpendicular to the velocity, the power developed by the force is zero. If power developed by a force is known it can be used to calculate the work done by that force. We have,

P = dw/dt ⇒

dw = P ⋅ dt t2

w = ∫ dw = ∫ P.dt t1

...(25)

Here w is the work done by the force which is developing the power P for the time interval [t1, t2 ]. Note: A common unit of power is horsepower. 1 horsepower = 746 watt

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Example – 13 A body of mass m is thrown at an angle α to the horizontal with an initial velocity v0 . Find the mean power developed by gravity over the whole time of motion of the body, and the instantaneous power of gravity as a function of time. Assume that the body is thrown at t = 0. Solution: The average power developed can be defined as the average rate of doing work. Therefore, the average power developed by gravity for the time interval [0, t] is .

P = =

∆wg ∆t ! ! Fg ⋅ ∆r ∆t

! ! = mg ⋅ v

! ∆r !    using ∆t = v 

= −mg ⋅ v y

 v y is the component of v!     in the upward direction

∆y ∆t

 ∆y is the displacement in the   vertical direction   

= −mg ⋅ = −mg

y(t ) − y(0) t −0

= −mg

y(t ) t

1 2   v0 sin α ⋅ t − gt  2  = −mg  t gt   = −mg  v0 ⋅ sin α −  2 

The average power developed for the entire duration of flight is zero, because for this duration ∆y is ! zero. Alternatively this could be explained in the following way: as the displacement, ∆r , is horizontal for this interval its scalar product with gravity is zero. The instantaneous power developed by gravity is ! ! ! ! P = Fg ⋅ v = mg ⋅ v = −mg ⋅ v y

!  v y is vertically upward component of v 

= −mg (v0 sin α − gt ) WORK, POWER & ENERGY

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From the equation above it is clear that while rising, the power developed by gravity is negative and while falling, the power developed by gravity is positive. (while rising v y is +ve and while falling v y is –ve.) Alternate Method:

We have ∆wg = −mg ∆h

Pg =

∆wg ∆t

=−

mg ⋅ ∆h ∆t

= −mg

∆h ∆t

Now, this result can be used to obtain the desired result.

Example – 14 For the situation given in example-10 find the dependence of the power developed by all forces on t and find the average value of this power for the first t seconds of motion. Solution: We have,

an = α t 2 ⇒

v2 = αt2 R

v = αR t

tangential acceleration, at =

at = α R

Normal component of net force,

dv dt

Fn = man = mα t 2 and tangential component of the net force,

Ft = mat = m α R.

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As we know that work is done by tangential component of the force, power is developed by this component only. We have, ! ! Pall = Fnet ⋅ v ! ! ! = ( Ft + Fn ) ⋅ v ! ! ! ! # v ∵ Ft || v and Fn ⊥ Pall = Ft ⋅ v ⇒ ...(26)

(

)

= m αR ⋅ αR t = mα Rt

Again, work done by all forces for the first t seconds of motion can be obtained by either finding the increment in kinetic energy of the particle or by integrating the power with respect to time. Therefore, for the time interval [0, t], work done by all forces is t

t

0

0

wall = ∫ Pall ⋅ dt = mα R∫ t ⋅ dt =

mα Rt 2 2

Therefore, average power developed by all forces for the same interval is

(

mα Rt 2 /2 work done Pav = = length of time interval (t ) =

Alternate-I:

)

mα Rt 2

We have, Pall = mα Rt , which is a linear function in t, therefore, the average value of Pall

Pav =

final value + initial value 2

Pav =

Pall (t ) + 0 2 =

mα Rt 2

Alternate-II: Using work energy theorem for the time interval [0, t] we get, wall = ∆k = k f − ki = k f 0

1 2 mα Rt 2 wall = mv = 2 2

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Average power developed by all forces =

work done by all forces lenth of time interval

=

(mα Rt 2/ 2) mα Rt = (t ) 2

Alternate-III: Average value of f ( x) over the interval [ x1, x2 ] is given as x2

f av =

∫ f (x) ⋅ dx

x1

[ x2 − x1]

t

pav = =

∫ p(t) ⋅ dt 0

(t − 0)

t

=

mα R∫ t ⋅ dt 0

t

mα Rt 2

Alternate-IV: We have tangential acceleration, at = α R , which is constant with time and we also have vinitial = 0, therefore, distance traveled by the particle along the path for the first t seconds is s=

1 1 ⋅ at ⋅ t 2 = α R t2 2 2

As the tangential force is constant and work done by normal force is always zero, work done by all forces for the same time interval can be written as

(

)

1  wall = Ft ⋅ s = m α R ⋅  ⋅ α R t 2  2  =

mα Rt 2 2

(mα Rt / 2) = 2

Therefore, Pall

(t )

=

mα Rt 2

You should try to understand the physical significance of the equation wall = Ft ⋅ s used here. Just give it a thought, and you are sure to get it.

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TRY YOURSELF - II Q. 1

A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown (a) vertically downward (b) vertically upward (c) horizontally (d) the speed does not depend on the initial direction.

Q. 2

Total work done on a particle is equal to the change in its kinetic energy (a) always (b) only if the forces acting on it are conservative (c) only if gravitational force alone acts on it (d) only if elastic force alone acts on it.

Q. 3

The kinetic energy of a particle continuously increases with time. (a) The resultant force on the particle must be parallel to the velocity at all instants (b) The resultant force on the particle must be at an angle less than 90° all the time (c) Its height above the ground level must continuously decrease (d) The magnitude of its normal acceleration is increasing continuously.

Q. 4

Under the action of a force, a 2 kg body moves such that its position x as a function of time t is given by t3 x = , x is in metre and t in second. Calculate the work done by the force in the first 2 second. 3

Q. 5

A block shown in figure slides on a semicircular frictionless track. If starts from rest at position A, what is its speed at the point marked B?

Q. 6

A 45º

1.0m B

An object of mass m is tied to a string of length l and a variable force F is applied on it which brings the string gradually at angle θ with the vertical. Find the work done by the force F.

θ l

m

Q. 7

A body of mass m accelerates uniformly from rest to v0 in time t0. As a function of t, the instantaneous power delivered to the body is : (a) m v / t (c) mvt 2 /t

Q. 8

F

(b) mv 2 /t (d) mv 2 /t 2 .

Show that for the same initial speed v0 the speed v of a projectile will be the same at all points at the same elevation, regardless of the angle of projection.

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Q. 9

31

A block of mass 10 kg slides down an incline 5 m. long and 3 m high. A man pushes up the block parallel to the incline so that it slides down at constant speed. The coefficient of friction between the block and the incline is 0.1. Find: (a) the work done by the man on the block (b) the work done by gravity on the block (c) the work done by the surface on the block (d) the work done by the resultant forces on the block (e) the change in K.E. of the block. [g = 10 m/s²]

Q. 10 A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation v = a x , where a is a constant. Find the total work done by all the forces during a displacement from x = 0 to x = d. Q. 11 Power delivered to a body varies as P = 3 t2. Find out the change in kinetic energy of the body from t = 2 to t = 4 sec. Q. 12 Figure shows a rough horizontal plane which ends in a vertical wall, to which a spring is connected, having a force constant k. Initially spring is in its relaxed state. A block of mass m starts with an initial velocity u towards the spring from a distance l0 from the end of spring, as shown. When block strikes at the end of the spring , it compresses the spring and comes to rest. Find the maximum compression in the spring. The friction coefficient between the block and the floor is µ . u m

k

µ l0

Q. 13 A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r as in figure. Measure angles from the vertical and potential energy from the top. Find (a) the change in potential energy of the mass with angle; (b) the kinetic energy as a function of angle; (c) the radial and tangential accelerations as a functions of angle (d) the angle at which the mass flies off the sphere. (e) If there is friction between the mass and the sphere, does the mass fly off at a greater or lesser angle than in part (d)? m θ

r

Q. 14 Two disks are connected by a stiff spring. Can one press the upper disk down enough so that when it is released it will spring back and raise the lower disk off the table (see figure)? Can mechanical energy be conserved in such a case? Q. 15 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is releases from there at zero speed with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides. WORK, POWER & ENERGY

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Section - 5

32

POTENTIAL ENERGY AND ENERGY CONSERVATION

While calculating work done by gravity and an ideal spring, you must have noticed that these were independent of the path of motion of the body. Work done by these forces depends only upon initial and final positions. The same idea leads to the fact that work done by these forces in a closed path is zero. In later chapters you would see that coulomb’s forces exhibit similar behaviour. This feature of these forces allow us to group them together in a different class of forces which are called conservative forces . Therefore a conservative force can be defined as a force whose work done is always independent of path. Alternatively, we can say that work done by a conservative force depends only upon initial and final positions or we can say that work done by a conservative force in a closed path is always zero. You can define a conservative force by any of the three statements. All these three statements have the same physical significance. Therefore if any force behaves like this, that force can be defined as a conservative force. Forces which do not satisfy these conditions can be defined as non conservative forces. For a while you are urged not to go into conceptual details of this new topic. We will first develop some methods based on the concept of conservative forces. Learn the method and its application initially, and then we will go into the conceptual details of conservative forces. The region in which a conservative force is acting is defined as the conservative force field of that force. For conservative force fields we associate potential energy with them. Why do we do so? This will be clear to you as you will proceed with this section. In a conservative force field work done by a conservative force is defined as negative of change in potential energy of the system. If potential energy is denoted by U, then, we define wcons = –∆U or ∆U = – wcon

...(27)

Therefore, for gravitational field near the earth’s surface, the change in gravitational potential energy, ∆U g = −wg = −(−mg ∆h)

[using (6)]

∆U g = +mg∆h

...(28)

and for a spring and block system,  1 1  ∆U sp = −wsp = − −  kx 2f − kxi2   2   2 ∆U sp =

1 2 1 2 kx f – kxi 2 2

[using (13)]

...(29)

Now, let us use the definition of change in potential energy in the work energy theorem. We have, wall = ∆k ⇒

wcon + wnoncon = ∆k

wnoncon = ∆k − wnoncon

wnoncon = ∆k + ∆U

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...(30) www.locuseducation.org

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Sum of kinetic and potential energies is defined as mechanical energy, E. Therefore, w noncon = ∆E

...(31)

i.e., work done by non conservative forces is equal to the change in mechanical energy of the system. Now, it should be clear that equations (30) or (31) are equivalent to work energy theorem. Instead of using work energy theorem we can use either of the two equations (30 or 31). So, what’s the advantage if we do this? The advantage is that if we proceed by this method we are not supposed to calculate the work done by the conservative forces. ∆U on the right side of the equation compensates for that. For any conservative force field generally we obtain a general expression for potential energy or change in potential energy. Once this result is available, we can use equation (30) for all motions in that field. We need not calculate the work done by conservative forces for all these motions separately. Equation (30) saves us from calculating work done by conservative forces which is a must if we follow the work energy theorem method. This is the basic idea behind the development of this concept. At this level you may not enjoy or realize the importance of this method because in problems at this level you can calculate both work done by conservative forces and change in potential energy very easily. But in the next level of study of science, in most of the cases you would find the potential energy method much simpler and easier than work energy theorem method, because calculating conservative forces and then their work done will be really cumbersome and more time consuming. You can consider the example of planetary motions and forces on molecular/atomic levels. Conservation of mechanical energy: If there are no non conservative forces present in a conservative force field or work done by non conservative forces is zero, then from equation (30) or (31), we have, ∆k + ∆U = 0

[when wnoncon = 0]

∆E = 0

or

...(32) ...(33)

i.e., change in mechanical energy of the system is zero. This is known as conservation of mechanical energy. From equation (32) it is clear that if ∆k is +ve then ∆U must be –ve and if ∆k is –ve then ∆U +ve, i.e., increase in kinetic energy is equal to decrease in potential energy and decrease in kinetic energy is equal to increase in potential energy. The total energy remains the same. If you want a change in the mechanical energy of a system, non conservative forces must do work on the system. Let us further extend equation (32). We have

∆k + ∆U = 0 ⇒

[when wnoncon = 0]

(k f − ki ) + (U f − U i ) = 0

k f + U f = ki + U i

E f = Ei

Note: While using change in potential energy or mechanical energy in work energy principle you must not consider work done by conservative forces. WORK, POWER & ENERGY

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Now, let us consider the equation (28) once again. We have, ∆U g = +mg ∆h

U f − U i = mg (h f − hi )

U f − U i = mgh f − mghi

...(34)

From the equation above we may infer that gravitational potential energy as a function of height (near earth surface) can be given as U (h) = mgh ...(35) But for the assumption above we need to clarify the following issues: 1.

If you assume U (h) = mgh + C , where C is a constant, then also equation (34) holds true;

2.

From where is the height ‘h’ measured

What is the reference level to measure the height h? The solution of both the problems lies in the fact that the potential energy is not defined absolutely. There is no absolute value of potential energy. We always define change in potential energy. If we insist on defining potential energy then any reference position can be defined as the zero potential energy configuration of the system. Consider equation (34) once again, we have, U f − U i = mgh f − mghi

U f − U ref = mgh f − mghref

If at h = href we define

 Reference level from which   height is measured is assigned     zero value, i.e., href = 0 

U f = mgh f

U = U ref = 0, then

U (h) = mgh

Similarly zero deformation (i.e., x = 0) is defined zero potential energy configuration for the spring and block system. In this case, we get U ( x) =

WORK, POWER & ENERGY

1 2 kx 2

[using equation (29)]

...(36)

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Example – 15 Solve example 2 using potential energy method. Solution: While the ball slides on the given surface only two forces act on it: (1) gravitational force; (2) normal contact force from the surface. As the gravitational force acting on the ball is conservative in nature, we will not consider the work done by it. Normal contact force acting on the ball is the only non conservative force acting on the ball. Therefore, from equation (30), we have

N h v smooth mg fig. 5.28

wnoncon = ∆U + ∆k

Alternate-I:

wN 0 = ∆U g + ∆k

∆U g + ∆k = 0

1  (mg ∆h) +  mv02 − 0  = 0 2 

1 mg (−h) + mv02 = 0 2

v0 = 2 gh

∵ N is always perpendicular   to the direction of motion   

v0 is the speed of the ball when it arrives  on the horizontal part of the surface   

Normal contact force acting the ball is the only non conservative force acting on it and work done by it zero, therefore, ∆U + ∆k = 0 ⇒

∆k = −∆U i.e., gain in kinetic energy = loss in gravitational potential energy ⇒

1 2 mv0 = mgh 2

v0 = 2 gh

Alternate-II: As the work done by non conservative forces is zero in this case, mechanical energy must be conserved. Therefore E f = Ei

U f + k f = U i + ki

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v=0

mgh f +

⇒ ⇒

1 2 1 mv f = mghi + mvi2 2 2 1 0 + mv02 = mgh + 0 2

36

h v0 smooth

mg

v0 = 2 gh initial position

final position

O P.E. level or O height level

fig. 5.29

Example – 16 Solve example 3 using potential energy method. Solution: Again this example can also be solved in various ways involving the concept of potential energy, as we did in the last example. As the ball slides down the curved part of the surface, its speed increases with distance. When the ball arrives on the horizontal part of the surface its speed becomes maximum. When the ball reaches the free end of the spring it begins to compress the spring and its speed decreases due to the force of the spring acting on it. When speed of the ball falls to zero, it can not compress the spring further. Hence when compression in spring is maximum, speed of the ball is zero. You should try to analyze the motion of the ball after this moment too. Therefore from the instant when the ball is released from rest to the moment when compression in the spring is maximum, there are three forces which acted on the ball: (1) weight of the ball, mg; (2) normal contact force, N; (3) spring force. Gravity and spring forces are conservative forces and normal contact force is the only non conservative force acting on the ball. As the normal contact force is always perpendicular to the movement of the ball, work done by it is zero. Method-I:

According to equation (30), we have wnoncon = ∆U + ∆k ⇒

wN = ∆U sp + ∆U g + ∆k

∆U sp + ∆U g + ∆k = 0

1 2 1 2 1 2 1 2  kx f − kxi  + [mg ∆h ] +  mv f − mvi  = 0 2 2 2  2 

1 2 kxmax − mgh + 0 = 0 2

xmax =

WORK, POWER & ENERGY

[∵

wN = 0]

 x f = xmax , xi = 0     ∆h = −h, vi = v f = 0 

2mgh . k www.locuseducation.org

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Method-II:

37

Here work done by non conservative forces is zero, therefore, total mechanical energy of the system remains the same. Hence, ∆U + ∆k = 0 ⇒

∆U sp + ∆U g + ∆k = 0

∆U sp + ∆U g = 0

∆U sp = −∆U g

gain in spring potential energy = loss in gravitational potential energy 1 2 kxmax = mgh 2 2mgh xmax = k

⇒ ⇒

∵ k f = ki = 0 

Method-III: As the total mechanical energy is conserved in this case, Ei = E f

U sp,i + U g ,i + ki = U sp, f + U g , f + k f

0 + mgh + 0 =

xmax =

1 2 kxmax + 0 + 0 2

∵

k f = ki = 0 

2mgh k

REST m

h xmax

k zero gravitational P.E. level

final position (b)

initial position (a) fig. 5.30

WORK, POWER & ENERGY

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Example – 17 1 m.

A string with one end fixed on a rigid wall passing over a fixed frictionless pulley at a distance of 2m from the wall has a mass M = 2 kg attached to it at a distance of 1 m from the wall. A mass m = 0.5 kg attached on the free end is held at rest so that the string is horizontal between the wall and the pulley and vertical beyond the pulley. What will be the speed with which the mass M will hit the wall when mass m is released? (g = 9.8 m/s²)

C

A B

m fig. 5.31

Solution: Here you should notice the following: •

1 m. M

Block B moves in a vertical circle with the centre at A when released from rest, as shown in figure 5.32. 1 m.

1 m. M

A

C

θ

B

1 m. u v

D θ

M

m h fig. 5.32

• •

If block M hits the wall horizontally with speed v, at that instant upward speed of block m is u = v cos θ . Gravity (a conservative force) is the only force doing net work on the system “block M + block m”, because work done by the part of the string between the block m and the vertical wall one the block m is zero (∵ tension force exerted by this part of the string is always perpendicular to the movement of the block) and the net work done by the remaining part of the string on the two blocks is also zero (although work done by it on a single block is nonzero) which can be proved as follows: C At any moment power delivered by the part of the string between the blocks to the block M is negative of power delivered by it to the block m. You can take the example of the moment just T u T D before block M hits the vertical wall, as shown v θ M in figure 5.33. At this moment if v be the speed m θ s o of the block M and u be the speed of the block vc m, as shown in figure, and T be the tension in the only tension forces on the thread, then power delivered to block m is Pm = +T ⋅ u

blocks from the part of the string between the blocks are shown fig. 5.33

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and that delivered to the block of mass M is Pm = −T ⋅ v cos θ Using constraint equation, we have,

u = v cos θ ∴

Method-I:

Pm = −PM

or we can say that the net power delivered to the system “block m and block M” by the string at this moment is zero. Similarly we can prove this fact for any moment. Hence net work done by the string on the two blocks during any time interval is zero. Alternate Way: As the string is massless, gain in its kinetic energy must be zero, therefore, net work done on it by the two blocks must be zero. Hence, net work done by the string on the two blocks is also zero. Now, let us solve for the required unknown v: Let us apply work energy theorem on the system “block m + block M” for the interval starting at the moment when the system was released from rest and ending at the moment when the block M is just about to hit the vertical wall. We have, wall = ∆k = k f − ki 1 1  wg + wT =  mu 2 + MV 2  − (0 + 0) 2 2  1 1 (+ Mg ⋅ AD − mg ⋅ h) + 0 = m(v cos θ ) 2 + MV 2 ⇒ 2 2 From figure 5.32, we have

...(i)

AD 1m 1 = = AC 2m 2 2 cos θ = ; ⇒ 5 and h = increase in the length of the string on the left side of the pulley tan θ =

= DC − BC

=

(

) (

22 + 12 − 1 =

)

5 −1

Putting these values in the equation (i), we get, 2 × 9 ⋅ 8 ×1 − 0 ⋅ 5 × 9 ⋅ 8 ×

(

)

1 4 1 5 − 1 = × 0 ⋅ 5 ×V 2 × + × 2 ×V 2 2 2 2 [it is given that m = 0.5 kg and M = 2.0 kg]

Solving the above equation we can find the value of v. WORK, POWER & ENERGY

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Method-II:

40

We have, wnoncon = ∆U + ∆k ⇒

wT = ∆U + ∆k

∆U + ∆k = 0

∆U m + ∆U M + ∆km + ∆k M = 0

1  1  (+mgh) + (−Mg.AD) +  mu 2 − 0  +  Mv 2 − 0  = 0 2  2 

[∵ wT = 0]

i.e., mechanical energy of the  system is conserved. Therefore,   we could also use   E = E ⇒ U + k =U + k  f i i f f   i

Now, putting the appropriate values as we did in the last method, we can solve for v.

Example – 18

VERTICAL CIRCULAR MOTION

In example 6 we have already calculated the speed of the ball and tension in the thread at some arbitrary angular position θ with respect to the lowermost position of the ball if the ball has a speed u at its lowermost position. But at that time you might have thought that what’s the possibility that the ball would complete the vertical circle at all?

O

m

u

fig. 5.34

Solution: If the ball is given a horizontal speed u when it is at its lowermost position, as shown in figure 5.34, the following three cases are possible: (a) The ball completes the vertical circle. (b) The ball does not reach the horizontal position and oscillates about its initial position, as shown in figure 5.35.

θ0 θ

Ball never reaches the horizontal position. It oscillates with angular amplitude θ about its vertical position.

θ0

fig. 5.35

(c) The ball succeeds in crossing the horizontal position but fails to complete the vertical circle. In this case we will also discuss whether, when the ball leaves the circle it has zero speed or zero tension in the string supporting it. Out of these three possible cases, what actually happens would depend upon the value of u. You must have an intuition that if u is very small then the ball would oscillate and if u is large than the ball would complete the vertical circle. WORK, POWER & ENERGY

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Now, consider the results obtained in example 6. At some arbitrary angular position θ speed of the ball, v, is given as v = u 2 − 2 gl (1 − cos θ )

v 2 = u 2 − 2 gl (1 − cos θ )

...(i)

and tension in the thread, T, is given as mu 2 T = mg (3cos θ − 2) + l m T = u 2 + gl (3cos θ − 2)  ⇒ l

...(ii)

CASE A: BALL COMPLETES THE VERTICAL CIRCLE: If the ball moves in a complete vertical circle, its distance from the point of suspension should be always equal to the length of the inextensible string supporting it and hence string should always be taut. Therefore, T>0 ⇒

m 2 u + gl (3cos θ − 2)  ≥ 0  l  2 ⇒ u + gl (3cos θ − 2) ≥ 0

u 2 ≥ gl (2 − 3cos θ )

u 2 ≥ maximum value of gl (2 − 3cos θ )

u 2 ≥ 5gl

[Using equation (ii)]

∵ at θ =π (topmost position)   ' 2-3cosθ ' has its maximum value 

u ≥ 5 gl

You should not that if u = 5gl , then at the topmost position T becomes zero but v is nonzero. For

u = 5gl at θ = π, equation (i) gives,

v 2/! mg

T=0 θ= π

v = gl and equation (i) gives, T=0

WORK, POWER & ENERGY

u = 5 gl fig. 5.36

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Therefore, for u = 5gl , at the topmost position gravity alone is providing the centripetal acceleration. This fact also gives mg = m ⇒

v2 l

v = gl

∴ velocity of the ball when it reaches the lowermost position is

u = 5gl

 using Ei = E f ⇒ k f = ki + loss in P.E.    1 2 1 2 ⇒ mu = mv + mg (2l) ⇒ u = 5gl    2 2

If the speed of the ball at the lowermost position, u, is greater than 5gl , then its speed at the topmost position is also greater than gl and hence more centripetal force is required. In this case both tension and gravity contribute to the centripetal force and hence T > 0. CASE B: BALL OSCILLATES WITH ANGULAR APTITUDE, θ0, SMALLER THAN π/2: Rearrange equations (i) and (ii) to get

and

v 2 = (u 2 − 2 gl ) + 2 gl cos θ

...(iii)

 l  T   = (u 2 − 2 gl ) + 3gl cos θ m

...(iv)

For this case we have θ ∈ [0, π /2). Therefore from equation (iii) and (iv) it is clear that if u 2 > 2 gl then neither v nor T becomes zero in this interval of θ. That is neither the ball stops nor the string becomes slack in this region. We can also say that if u 2 > 2 gl then string would definitely cross the horizontal position (i.e., θ = π/2 position). Therefore if u 2 < 2 gl then the ball can not deflect by π/2 and it would oscillate about its lower most position with angular amplitude smaller than π/2. In such a case, it is clear from equations (iii) and (iv) that speed of the ball vanishes before the tension in the string. That is, in such a case tension in the string is never zero and speed of the ball is zero at the extreme position (the position from where it starts returning back towards lowermost position). These facts could also be explained in the following way: The position of the ball is shown at some arbitrary angular position θ in figure 5.37(a). If v be the speed of the ball at this position then it is obvious from the figure that at this moment tension force is balancing the radially outward component of gravity as well as providing the required centripetal acceleration to the ball and hence it is greater than mg cos θ (radially outward component of gravity). Here tangential component of gravity, mg sin θ , is retarding the upward (along the circle) motion of the ball as shown in figure 5.37(a). When speed of the ball becomes zero at some angle θ 0 , then also tension has to balance the radially outward component of gravity, mg cos θ 0 , WORK, POWER & ENERGY

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and hence it can not be zero, as shown in figure 5.37(b). In the same figure you should notice that, at θ = θ0, mg sin θ would accelerate the ball towards its lower most position.

θ0

θ

T

T

v=0

v gs

mg co sθ

0

m

cos

inθ

mg

s mg

0

in

θ

v !

2

θ

fig. 5.37 (a) (T > mg cos θ)

fig. 5.37 (b) (T = mg cos θ )

If u 2 = 2 gl then according to equations (iii) and (iv), v and T both vanish simultaneously at θ = π/2, as shown in figure 5.38. In this case the ball oscillates with angular amplitude π/2. T=0 fig. 5.38 (ball just reaches the horizontal position)

v=0

π/2 mg u = 2gl

CASE C: BALL CROSSES THE HORIZONTAL POSITION BUT DOES NOT COMPLETE THE VERTICAL CIRCLE. From the previous two cases it must be clear to you that if u is greater than 2gl but smaller than 5gl then the ball would deflect more than π /2 but it can not complete the vertical circle. Therefore, it would leave the circular path for an angle α greater than π/2 but smaller than π. From equations (iii) and (iv) it can be concluded that if u 2 > 2 gl and θ > π /2 then T vanishes before v (only if u 2 < 5gl ), i.e., at some angle α (as mentioned above) the string becomes slack but the ball still has some nonzero speed, as shown in figure 5.39. After this particular position gravity is the only force acting on the moving ball and it has already left the circular path (because the slack string means that the distance of the ball from the point of suspension is smaller than the length of the string), therefore, motion of the ball would be equivalent to that of a projectile moving in a parabolic path.

WORK, POWER & ENERGY

v T=0

α

mg

u fig. 5.39

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ALTERNATE APPROACH: Let us divide the vertical circle under consideration in four quadrants, as shown in figure 5.40. Now, we will analyze the each quadrant separately.

Vertical line

III

II

IV

I

Horizontal line

fig. 5.40

θ v2 !

mg

sin

θ

θ

fig. 5.41

v cos

u

T

mg

Quadrant I: Some arbitrary position of the ball in this quadrant is shown in figure 5.41. From this figure it is clear that the radial component of the gravity is in the opposite direction of the radial acceleration of the ball and hence in this quadrant tension has two roles: first is to balance the radial component of gravity and second is to provide the necessary centripetal acceleration to the ball, as discussed before. Hence, in this quadrant tension can never be zero.

If the ball just manages to reach the horizontal position then at this moment alongwith the speed of the ball tension in the thread also becomes zero because at this position the radial acceleration and the radial component of gravity both are zero, as shown in figure 5.38 and hence there is no requirement of tension in the thread. Using work energy theorem or conservation of mechanical energy we can prove that for this to happen u should be 2gl . Quadrant II: Some arbitrary position of the ball when it is in quadrant II is shown in figure 5.42(a). It is obvious from the figure that as the ball moves up, its speed decreases and component of gravity along the radially inward direction increases and hence requirement of tension force becomes less and less as the ball moves up in this quadrant. Eventually at the highest point of the circle the speed of the ball becomes minimum and the contribution of gravity in centripetal force becomes maximum and hence at this position requirement of tension is minimum, as shown in figure 5.42(b). Therefore this position can be defined as the critical position, because if the ball crosses this position successfully, i.e., if the string is taut in this position, it would always be taut or we can say that the ball would complete the vertical circle. v

m gs in θ

θ

fig. 5.42 (a)

WORK, POWER & ENERGY

θ os gc

O

T

m

v 2/!

v T

mg v 2/!

O

fig. 5.42 (b) (requirement of tension is minimum at this position, i.e., the chance of slacking of the string is maximum at this position) www.locuseducation.org

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If the ball is just completing the vertical circle we can assume zero tension at the critical position (topmost point) because gravity is there to provide the required centripetal force to the ball. In such a case if v be the speed of the ball at the highest position, then mg = ⇒

mv 2 l

v 2 = gl

and if u be the speed of the ball when it reaches the lowermost position, then using k f = ki + loss in P.E.

we get, 1 1 mu 2 = mv 2 + mg (2l ) 2 2

u 2 = gl + 4 gl

u = 5gl

Hence, to complete the vertical circle, u ≥ 5 gl . Note: •

To just complete the vertical circle you can not assume zero speed at the highest point. Why so? Try to answer it on your own.

The only difference between the analysis of the ball in quadrant II and III is that when the ball is moving in quadrant II, it is speeding down and while it is moving in quadrant III, it is speeding up. Similar argument can be given for quadrants I and IV.

Summary: •

When u ≤ 2 gl

:

The ball oscillates with angular amplitude, θ0 ≤ π /2.

When 2 gl < u < 5 gl

:

The ball leaves the vertical circular path at some position (which would depend upon u) in the IInd quadrant and thereafter moves in a parabolic path.

When u ≥ 5 gl

WORK, POWER & ENERGY

:

The ball moves in a complete vertical circular path.

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TRY YOURSELF - III Q. 1

Q. 2

Q. 3

An object of mass m is tied to a string of length l and a variable force F is applied on it which brings the string gradually at angle θ with the vertical. Find the work done by the force F. [Solving using potential energy method] A body is dropped from a certain height. When it lost an amount of P.E. ‘U’, it acquires a velocity ‘v’. The mass of the body is : (b) 2v / U 2 (c) 2v/U (a) 2U / v ²

θ l

F

m

(d) U 2/2v.

A particle of mass m is attached to a light string of length l, the other end is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle. (a) The string becomes slack when the particle reaches its highest point (b) The velocity of the particle becomes zero at the highest point 1 2 (c) The kinetic energy of the ball in initial position was mv = mgl. 2 (d) The particle again passes through the initial position.

Q. 4

Q. 5

A small block of mass m slides along the frictionless loop-theloop track shown in fig. (a) If it starts from rest at P, what is the resultant force acting on it at Q? (b) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop is equal to its weight? A simple pendulum of length l, the mass of whose bob is m, is observed to have a speed v0 when the cord makes the angle θ0 with the vertical (0 < θ 0 < π /2), as in fig. In terms of g and the foregoing given quantities, determine (a) the speed v1 of the bob when it is at its lowest position; (b) the least value v2 that v0 could have if the cord is to achieve a horizontal position during the motion.

P 5R

R

Q

θ0 l

m v0

Q. 6 A chain of length l and mass m lies on the surface of a smooth hemisphere of radius R > l with one end tied to the top of the hemisphere. Find the gravitational potential energy of the chain with respect to the given P.E. level.

R ZERO P. E. LEVEL

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Q. 7

A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero speed with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides. [Solve using potential energy method]

Q. 8

A block rests on an inclined plane as shown in figure. A spring to which it is attached via a pulley is being pulled downward with gradually increasing force. The value of µ s is known. Find the potential energy U of the spring at the moment when the block begins to move.

m

µs θ

Q. 9

k F

The particle m in figure is moving in a vertical circle of radius R inside a track. There is no friction. When m is at its lowest position, its speed is v0 . (a) What is the minimum value vm to v0 for which m will go completely around the circle without losing contact with the track? (b) Suppose v0 is 0.775 vm. The particle will move up the track to some point at P at which it will lose contact with the track and travel along a path shown roughly by the dashed line. Find the angular position θ of point P.

P θ

R v0 m

Q. 10 A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere. (a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere. (b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slide through an angle θ. dv (c) Find the tangential acceleration of the chain when the chain starts sliding down. dt Q. 11 A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure). The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ (shown in the figure)

Sphere B θ d

O

R Sphere A

(a) Express the total normal reaction forces exerted by the spheres on the ball as a function of angle θ. (b) Let N A and N B denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and B, respectively. Sketch the variations of N A and N B as functions of cos θ in the range 0 ≤ θ ≤ π by drawing two separate graphs, taking cos θ on the horizontal axes. Also sketch the variations of NA and NB as functions of θ. WORK, POWER & ENERGY

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Section - 6

48

MORE ON CONSERVATIVE FORCES

Conservative forces and potential energy again The interaction of a particle with surrounding bodies can be described in two ways : by means of forces or through the use of the notion of potential energy. In classical mechanics both ways are extensively used. The first approach, however, is more general because of its applicability to forces in the case of which the potential energy is impossible to introduce (i.e., non conservative forces). As to the second method, it can be utilized only in the case of conservative forces. Our objective is to establish the relationship between potential energy and the force of the conservative field, or ! ! ! putting it more precisely, to define the conservative field of forces F (r ) from a given potential energy U (r ) as a function of a position of a particle in the field. We have learnt by now that the work performed by conservative forces on a particle during the displacement of the particle from one point in the field to another may be described as the decrease of the potential energy of the particle, that is, wcon = −∆U . The same can be said about the elementary displacement dr! as well:

or

dwcon = −dU ! ! F ⋅ dr = –dU

...(37)

! ! ! Recalling that dw = F ⋅ dr = Ft ⋅ ds, where ds = dr is the elementary length covered along the path and Ft is ! the tangential component of F , we shall rewrite equation (37) as Ft ⋅ ds = −dU .

Hence,

Ft = −

∂U ∂s

...(38)

! i.e., the projection of the conservative force at a given point in the direction of the displacement dr equals the derivative of the potential energy U with respect to a given direction, taken with the opposite sign. The designation of a partial derivative ∂/∂s emphasizes the fact of differentiating with respect to a finite direction. ! The displacement dr can be resolved along any direction and, specifically, along the x, y, z coordinate axes. For ! ! example, if displacement dr is parallel to the x axis, it may be described as dr = dxiˆ. The work performed by ! the conservative force F over the displacement dr! parallel to the x axis is ! ! ! F ⋅ dr = F ⋅ (dxiˆ) = Fx ⋅ dx, ! where Fx is the x-component of the force F . Substituting the last expression into equation (38), we get Fx = −

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∂U ∂x

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where the partial derivative symbol implies that in the process of differentiating U ( x, y, z) should be considered as a function of only one variable, x, while all other variables are assumed constant. It is obvious that the equations for Fy and Fz are similar to that for Fx. So, having reversed the sign of the partial derivatives of the function U ! with respect to x, y, z, we obtain the components Fx , Fy and Fz of the conservative force F . Hence, we have

! F = Fxiˆ + Fy ˆj + F2kˆ ! ∂U ˆ ∂U ˆ ∂U ˆ F = – i+ j+ k ∂y ∂z  ∂x

...(40)

The quantity in parentheses is referred to as the scalar gradient of the function U and is denoted by grad U or ∇U. Generally the second, more convenient, designation where ∇(“nabla”) signifies the operator ∂ ∂ ˆ ∂ ˆ ∇ = iˆ + j+ k ∂x ∂y ∂z is used. Consequently, we can write,

! F = − ∇U

..(41)

! i.e., the conservative!force F is equal to the potential energy gradient, taken with the minus sign. Put simply, the conservative force F is equal to the antigradient of potential energy.

Example – 19 The potential energy of a particle in certain conservative field has the following form: (a) U(x, y) = – αxy, where α is a constant; ! ! ! ! ! (b) U (r ) = a ⋅ r , where a is a constant vector and r is the position vector of the particle in the field. Find the conservative field force corresponding to each of these cases. Solution: (a) We have, !  ∂U ˆ ∂U ˆ  F = − i+ j ∂y   ∂x =

∂(α xy) ˆ ∂(α xy) ˆ i+ j ∂x ∂y

=αy

∂x ˆ ∂y ˆ i +α x j ∂x ∂y

= α ( yiˆ + xjˆ)

(b) We have,

! a = a xiˆ + a y ˆj + az kˆ and ! r = xiˆ + yjˆ + zkˆ

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Therefore,

50

! ! U = a⋅r = a x ⋅ x + a y ⋅ y + a z ⋅ z; then

!  ∂U ˆ ∂U ˆ ∂U F = − i+ j+ ∂y ∂y  ∂x = −(axiˆ + a y ˆj + az kˆ)

 kˆ  

! = −a

Example – 20 A conservative force F(x) acts on a 1.0 kg particle that moves along the x-axis. The potential energy U(x) is given as U ( x) = a + ( x − b) 2 where a = 20 J, b = 2 m. and x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J. It is known that there is no other force acting on the system. Based upon this information, answer the following questions: (a) What is the mechanical energy of the system? (b) What is the range of x in which the particle can move? (c) What is the maximum kinetic energy of the particle and the position where it occurs? (d) What is the equilibrium position of the particle? Solution: We have, U ( x) = a + ( x − b) 2 = 20 + ( x − 2) 2

We know that at x = 5.0, K.E. = 20 J, therefore, mechanical energy of the particle, E = Potential energy, U + kinetic energy, k = U (at x = 5) + k (at x = 5)

= 20 + (5 − 2) 2 + 20 = 49 J .

[Ans.(a)]

We know that, E= U+k

k = E −U

= 49 −  20 + ( x − 2) 2  = 29 − ( x − 2) 2 WORK, POWER & ENERGY

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As the particle moves on the x-axis, its kinetic energy can never be negative, therefore, we have,

k ≥0 ⇒

29 − ( x − 2) 2 ≥ 0

( x − 2) 2 ≤ 29

x 2 − 4x + 4 − 29 ≤ 0

x 2 − 4 x − 25 ≤ 0

x ∈ [−3.38, 7.38]

[Ans (b)]

When the particle has maximum kinetic energy, we have, dk =0 dx ⇒

d  29 − ( x − 2) 2 

0 − 2 ⋅ ( x − 2) ⋅ (1) = 0

x=2

dx

=0

i.e., at x = 2 , the particle has maximum kinetic energy which is equal to 29 J.

[Ans. (c)]

As the conservative force is antigradient of potential energy, we have, F ( x) = −

=−

dU dx

d  20 + ( x − 2)2  dx

= − [0 + 2 ⋅ ( x − 2) ⋅ (1) ] = 4 − 2x .

When the particle is in equilibrium, net force on it must be zero. As the only force acting on the particle is F(x), in equilibrium position F ( x) = 0

4 − 2x = 0

x = 2m.

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[Ans.(d)]

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ALTERNATE METHOD: Let us solve this problem graphically. For that first calculate mechanical energy, E. Following the same procedure as we did in the last method, we get E = 49J. and we already have, U ( x) = 20 + ( x − 2) 2. The plots of E and U(x) are shown in figure 5.43. From the graph it is clear that at x = 7.38 and x = –3.38 U becomes equal to E, hence, at these positions k = 0 (because E = k + U). For x > 7.38 and x < –3.38, U become greater than E and hence k would acquire –ve value, which is never possible. Therefore, particle can move only for x greater than –3.38 and smaller than 7.38. As the sum of k and U is constant, k would be maximum when U is minimum. Therefore, at x =2, k is maximum and is equal to 49 – U = 49 – 20 = 29 J. At x = 2, slope of U is zero, therefore, at x = 2, F(x), which negative of slope of U, is also zero. Therefore, x = 2 is equilibrium position of the particle.

U (J ) E =49J 49

20 –3.38 O

2

X(m)

7.38

fig. 5.43

U

Nature of Equilibrium: Whenever a conservative force is the only force acting on a particle, equilibrium positions of the particle can be determined from the graph of U when plotted against the position of the particle. Consider the case shown in figure 5.44. In this figure potential energy, U, of a particle under the action of a conservative force

O

fig. 5.44

F(x) is plotted against the position of the particle, x. It is obvious from the graph of U that at x = x1 it has a maxima and at x = x2 it has a minima. Therefore, at both x1 and x2 derivative of U is zero and hence F(x) is zero. Consequently x1 and x2 are equilibrium positions of the particle. Therefore, we can say that extrema of U occur at equilibrium positions of the particle.

x

x2

x1

U

Now, let us analyze the force on the particle when it is in the vicinity of one of its equilibrium positions or it is in the vicinity of an extre ma of its potential energy. For values of x very close to x1 but smaller than x1 derivative of U is positive because its tangent makes an acute angle with the +ve direction of the x-axis. Therefore, in this region the force on the particle is along negative x direction (or we can say that it is away from x1 ), as shown in figure 5.45. Similarly, it can be WORK, POWER & ENERGY

O

F (x )

x1

x F (x )

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justified that for values of x very close to x1 but greater than x1 force, F(x), is positive, i.e., it acts away from x1 , as shown in figure 5.45. Now, suppose a particle in equilibrium at x = x1 is slightly displaced from this position the either side and released under the action of conservative force, F(x), only. What would happen now? The force on the particle would act away from the equilibrium position x = x1 and the particle would never come back to this equilibrium position. Such an equilibrium position is defined as unstable equilibrium position. Now, let us analyze the behaviour of F(x) in the vicinity of a minima of U. In figure 5.46 x = x2 is a minima of U. Arguments similar to what I provided in the previous paragraph lead to the fact that in the vicinity of x2 (a minima of potential energy) conservative force, F(x), acts always towards x2 . Therefore, if a particle is displaced from x2 on either side and released from rest and thereafter only conservative force, F(x), acts upon it, it will return back to its equilibrium position ( x = x2 ) . (A little more thought would give an idea of oscillation about equilibrium position, which I don’t want to discuss here.) Such an equilibrium position is defined as a unstable equilibrium position.

y

o

F (x )

x2

x F (x )

fig 5.46

Therefore, the position where potential energy, U, Possesses a maxima (first derivative of U is zero and second derivative is negative) is an unstable equilibrium position and the position where potential energy possesses a minima (first derivative of U is zero and second derivative is positive) is a stable equilibrium position. Therefore, in example 20, x = 2, was a stable equilibrium position.

Example – 21 The potential energy of a particle in a certain field has the form U = a/r 2 − b/r, where a and b are positive constants, r is the distance from the centre of the field. Find: (a) the value of r0 corresponding to the equilibrium position of the particle, examine whether this position is stable; (b) the range of the attraction force. Solution: (a) At equilibrium position: dU =0 dr d (a/r 2 − b/ r ) =0 ⇒ dr ⇒

−2a b + =0 r3 r2

b=

2a r

r=

2a b

r0 = 2a/b

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Now,

d U d ( − dU dr ) = = dr dr 2 2

=

54

 2a b  d − 3 + 2  r   r dr

6a 2b − dr 4 r 3

d 2U 6a 2b = − dr 2 r04 r03

At r = r0 ,

6a ⋅ b4 2b ⋅ b3 = − 16a 4 8a3 3 b4 1 b4 = − 8 a3 4 a3 = Therefore, at r = r0 ,

1 b4 8 a3

d 2U is positive and hence this is a stable equilibrium position. dr 2

(b) Radial component of force, Fr (r ), is antigradient of U with respect to r. therefore, Fr (r ) = − =

dU dr

2a b − r3 r2

When this force is attractive, it must be along radically inward direction and hence it should be negative. That is,

Fr (r ) < 0 ⇒

2a b −

2a 2a/b, Fr (r ) is , Fr (r ) is repulsive, at r = b b

attractive.

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Section - 7

55

A NOTE ON MASS AND ENERGY (OPTIONAL)

Mass and Energy One of the great conservation laws of science has been the law of conservation of matter. From a philosophical point of view an early statement of this general principle was given by the Roman poet Lucretius, a contemporary of Julius Caesar, in his celebrated work De Rerum Natura. Lucretius wrote “ Things cannot be born from nothing, cannot when begotten be brought back to nothing.” It was a long time before this concept was established as a firm scientific principle. The principal experimental contribution was made by Antoine Lavoisier (17431794), regarded by many as the father of modern chemistry. He wrote in 1789 “We must lay it down as in incontestable axiom, that in all the operations of art and nature, nothing is created; and equal quantity of matter exists both before and after the experiment .... and nothing takes place beyond changes and modifications in the combinations of these elements.: This principle, subsequently called the conservation of mass, proved extremely fruitful in chemistry and physics. Serious doubts as to the general validity of this principle were raised by Albert Einstein in his papers introducing the theory of relativity. Subsequent experiments on fast moving electrons and on nuclear matter confirmed his conclusions. Einstein’s findings suggested that, if certain physical laws were to be retained, the mass of a particle had to be redefined as m0 m= ...(42) 1 − v 2 /c 2 . Here m0 is the mass of the particle when at rest with respect to the observer, called the rest mass; m is the mass of the particle measured as it moves at a speed v relative to the observer; and c is the speed of light, having a constant value of approximately 3 ×108 meters/sec. Experimental checks of this equation can be made, for example, by deflecting high-speed electrons in magnetic fields and measuring the radii of –31

18×10

–31

16×10 m, kg

–31

14×10

fig 5.47

[ The way an electron’s mass increases as its speed relative to the observer increases. The solid line is a

–31

plot of m= m0 (1 − v 2/c 2 ) −1/2, and the circles are

–31

adapted from experimental values. The curve tends toward infinity as v → c. ]

12×10 10×10

0

0.2

0.4

0.6 v/c

0.8

1.0

curvature of their path. The paths are circular and the magnetic force a centripetal one ( F = mv 2 /r , F and v being known). At ordinary speeds the difference between m and m0 is too small to be detectable. Electrons, however, can be emitted from radioactive nuclei with speeds greater than nine-tenths that of light. In such cases the results (figure) confirm Equation (42) It is convenient to let the ratio v/c be represented by β. The equation (42) becomes

m = m0 (1 − β 2 )−1/2. To find the kinetic energy of a body, we compute the work done by the resultant force in setting the body in motion. We have, WORK, POWER & ENERGY

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v ! ! 1 K = ∫ F ⋅ dr = m0v 2 0 2

for kinetic energy, when we assumed a constant mass m0. Suppose now instead we take into account the variation of mass with speed and use m = m0 (1 − β 2 )−1/2 in our previous equation. We find that the kinetic energy is no 1

longer given by 2 m0v 2 but instead is

K = mc 2 − m0c 2 = (m − m0 )c 2 = ∆mc 2.

...(43)

2 The kinetic energy of a particle is, therefore, the product of c and the increase in mass ∆m resulting from the motion.

Now, at small speeds we expect the relativistic result to agree with the classical result. By the binomial theorem we can expand (1 − β 2 ) −1/2 as (1 − β 2 ) −1/2 = 1 + β 2 + β 4 + 1 2

3 8

5 16

β 6 + ........ .

At small speeds β = v / c % 1 so that all terms beyond β 2 are negligible. Then K = (m − m0 )c 2 = m0c 2 (1 − β 2 ) −1/2 − 1

(

1 2

)

1 2

1 2

= m0c 2 1 + β 2 + ....... − 1 ≅ m0c 2 β 2 = m0v 2 ,

which is the classical result. Notice also that when K equals zero, m = m0 as expected. The basic idea that energy is equivalent to mass can be extended to include energies other than kinetic. For example, when we compress a spring and give it elastic potential energy U, its mass increases from m0 to

m0 + U/c 2. When we add heat in amount Q to an object, its mass increases by an amount ∆m, where ∆m is Q/c 2. We arrive at a principle of equivalence of mass and energy: For every unit of energy E of any kind supplied to a material object, the mass of the object increases by an amount ∆m = E/c 2 This is the famous Einstein formula

E = ∆mc 2 .

...(44)

In fact, since mass itself is just one form of energy, we can now assert that a body at rest has an energy m0c 2 by virtue of its rest mass. This is called its rest energy. If we now consider a closed system, the principle of the conservation of energy, as generalized by Einstein, becomes

∑(m0c 2 + ∈) = constant or

∆(∑ m0c 2 + ∑ ∈) = 0,

2 where ∑ m0c is the total rest energy and ∑ ∈ is the total energy of all other kinds. As Einstein wrote, “Prerelativity physics contains two conservation laws of fundamental importance, namely the law of conservation of energy and the law of conservation of mass; these two appear there as completely independent of each other. Through relativity theory they melt together into one principle.”

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Because the factor c 2 is so large, we would not expect to be able to detect changes in mass in ordinary mechanical experiments. A change in mass of 1 gm would require an energy of 9 ×1013 joules. But when the mass of a particle is quite small to begin with and high energies can be imparted to it, the relative change in mass may be readily notice able. This is true in nuclear phenomena, and it is in this realm that classical mechanics breaks down and relativistic mechanics receives its most striking verification. A beautiful example of exchange of energy between mass and other forms is given by the phenomenon of pair annihilation or pair production. In this phenomenon an electron and a positron, elementary material particles differing only in the sign of their electric charge, can combine to literally disappear. In their place we find highenergy radiation, called γ-radiation, whose radiant energy is exactly equal to the rest mass plus kinetic energies of the disappearing particles. The process is reversible, so that a materialization of mass from radiant energy can occur when a high enough energy γ-ray, under proper conditions, disappears; in its place appears a positronelectron pair whose total energy (rest mass + kinetic) is equal to the radiant energy lost.

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TRY YOURSELF - IV

Q. 1

Show that mc2 has the dimensions of energy.

Q. 2

If the potential energy of a two particle system separated by a distance r is given by U (r ) =

A , where A r

is a constant, find the redial force, Fr , that each particle exerts on the other. Q. 3

The potential energy of a conservative system is given by U = ax 2 − bx where a and b are positive constants. find the equilibrium position and discuss whether the equilibrium is stable or unstable.

Q. 4

Give physical examples of unstable equilibrium. Of stable equilibrium.

Q. 5

“The position of maximum potential energy is the position of unstable equilibrium”. State whether the statement is true of false, with short reason.

Q. 6

For the potential energy curve shown in figure (a)

(b)

U

Determine whether the conservative force F is positive, negative or zero at the five points A, B, C, D and E. Indicate points of stable and unstable equilibrium.

B C

A

2

6

8

4 D

x(m)

E

U

Q. 7

In the figure shown the potential energy U of a particle is plotted against its position ‘x’ from origin. Then which of the following statement is correct. A particle at : (a) (b) (c) (d)

Q. 8

x1 is in stable equilibrium x2 is in stable equilibrium x3 is in stable equilibrium none of these

O

The given plot shows the variation of U, the potential energy of interaction between two particles with the distance separating them, r :

U

x1

x2

x3

x

A E

(i) (ii) (iii)

B and D are equilibrium points F B C is a point of stable equilibrium r D The force of interaction between the two particles C is attractive between points C and D and repulsive between points D and E on the curve. (iv) The force of interaction between the particles is repulsive between point E and Fon the curve. Which of the above statements are correct. (a) (i) and (iii) (c) (ii) and (iv)

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(b) (d)

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Q. 9

59

A certain peculiar spring is found not to confirm to Hooke’s law. The force (in Newton’s) it exerts when stretched a distance x (in meters) is found to have magnitude 52.8x + 38.4x 2 in the direction opposing the stretch. (a) Compute the total work required to stretch the spring from x = 0.50 to x = 1.00 meter. (b) With one end of the spring fixed, a particle of mass 2.17 kg is attached to the other end of the spring when it is extended by an amount x = 1.00 meter. If the particle is then released from rest, compute its speed at the instant the spring has returned to the configuration in which the extension is x = 0.50 meter. (c) Is the force exerted by the spring conservative or non conservative? Explain.

Q. 10 If the magnitude of the force of attraction between a particle of mass m1 and a mass m2 is given by mm F = k 1 2 2 where k is a constant and x is the distance between the particles, find (a) the potential x energy function and (b) the external work required to increase the separation of the masses from x = x1 to x = x1 + d. [Assume, U = 0 when x → ∞ ]. Q. 11 The magnitude of the force of attraction between the positively charged nucleus and the negatively charged e2 electron in the hydrogen atom is given by F = k 2 where e is the charge of the electron, k is a constant, r and r is the separation between electron and nucleus. Assume that the nucleus is fixed. The electron, initially moving in a circle of radius R1 about the nucleus, jumps suddenly into a circular orbit of smaller radius R2. (a) (b)

Calculate the change in kinetic energy of the electron, using Newton’s second law. Using the relation between force and potential energy, calculate the change in potential energy of the atom. (c) Show by how much the total energy of the atom has changed in this process. (The total energy will prove to have decreased; this energy is given off in the form of radiation.) [Assume, U = 0 when x → ∞ ]. Q. 12 Given below (figure) are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

U (x )

U3 U2 U1

(a) U0 E

(b) U0

(c)

E

O

a

x

O

U(x)

U(x)

U(x)

a b c d

x

U0

U0 E O –U1

E

a

b

x

(d)

-b/2 -a/2

a/2 -b/2

x

U1

.

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Section - 8

60

MISCELLANEOUS EXAMPLES

Example – 22 A block of mass 2.0 kg is moved from rest on a rough horizontal surface by applying a force F ( x) = 15 + x − x 2 from x = 0 ⋅ 0 to x = 1⋅ 0. If friction coefficient between the block and horizontal surface be 0 ⋅ 2 , find the gain in kinetic energy of the block. [Take g = 10 m/s²] Solution: According to the work energy theorem,

∆k = work done by all forces = work done by F ( x) + work done by friction  as the block is moving on a horizontal surface,  work done by gravity and normal contact   force from the surface is zero

=

x2

∫ F ( x) ⋅ dx − µ mg ⋅ d

   

[d = 1 ⋅ 0 m.]

x1 1

= ∫ (15 + x − x 2 ) ⋅ dx − 0 ⋅ 2 × 2 ⋅ 0 × 10 × 1 ⋅ 0 0

1

= 15

1 x0

1

x2 x3 + − −4 2 0 3 0

1 1   =  15 + − − 4  2 3   = 11.17 J.

Example – 23 A body of mass m! was slowly hauled up the hill shown in figure 5.48 by a face F which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is h, the length of its base l, and the coefficient of friction µ. Solution: An elementary part, of length ds, of the path of the body and forces acting upon the body when it was on that part of the path are shown in figure 5.49. The applied external force F is acting tangentially up the surface, friction force f is acting tangentially down WORK, POWER & ENERGY

! F m h

l fig. 5.48

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the surface, normal contact force N is perpendicular to the surface (and hence is perpendicular to the direction of motion of the body) and weight, mg , of the body is acting vertically downwards. F N

ds θ

ds m

θ

f

F

N

f

mg cosθ

mgsinθ mg fig. 5.49

As the body is being moved slowly, net force on the body should be zero and friction would be kinetic in nature. Therefore, F = µ N = µ mg cos θ F = f + mg sin θ = µ mg cos θ + mg sin θ

and

As the body is moved over this elementary part of the path, from figure 4.9, work done by F is dw = F ⋅ ds

[∵ ds is along F ]

= [µ mg cos θ + mg sin θ ] ⋅ ds = mg[µ ⋅ ds ⋅ cos θ + ds ⋅ sin θ ]

Now, if we define the horizontal direction as the x-direction and vertically upward direction as the ydirection, as shown in figure 5.50, then we have

ds ⋅ cos θ = dx and

y

ds ⋅ sin θ = dy

ds dy θ dx

Therefore,

x

dw = µ mg ⋅ dx + mg ⋅ dy

fig. 5.50

Hence, net work done by force F is w = ∫ dw =

fin

µ mg ⋅ dx +

in

fin

∫ mg.dy

in

w = µmg ⋅ l + mg ⋅ h

Note: As the body is moved slowly or we can say as the equilibrium is always maintained, gain in K.E. of the body must be zero. Therefore, using work energy theorem, we get, WORK, POWER & ENERGY

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wall = ∆k = 0 ⇒

wF + wN + w f + wg = 0

(µ mgl + mgh) + 0 + w f − mgh = 0

w f = −µ mgl

work done by friction force = –µmg • l

Now, notice the result obtained above for the work done by the friction force. This work done is equivalent to the work friction would have done if the body was given the same horizontal displacement on a horizontal surface with the same roughness.

Example – 24 AB is a quarter of a smooth circular track of radius r = 4m., as shown in figure 5.51. A particle P of mass m = 5 kg moves along the track from A to B under the action of the following forces: (i)

A force F1, always directed towards B, its magnitude is constant and is equal to 4 newton.

(ii)

A force F2, which always directed along the tangent to the circular track, its magnitude is (20 – s) newton, where s is the distance traveled.

(iii)

A constant horizontal force F3 of magnitude 25 newton.

If the particle starts at A with a speed of 10 m/s, what is the speed at B?

O

r

B F1

r

F2 F3

A fig. 5.51 r

B Solution: First of all let us calculate work done by each of the O π 2 θ three forces. ! θ F1 Work done by F1 and F2 : When the particle has already ! dl r α moved a distance s away from A its position is shown in figure 5.52. If the particle has rotated about the centre of the P circular path, point O, by an angle θ and α be the angle ! s A between OB and direction of F1, then in ∆OPB, we have, π  fig. 5.52  − θ  + (α ) + (α ) = π 2  π 2α = + θ ⇒ 2 π θ α= + ⇒ 4 2 ! ! Now, if the particle is displaced by dl in next time interval dt, work done by F1 is ! ! dw1 = F1 ⋅ d l

π  = F1 ⋅ dl ⋅ cos  − α  2  WORK, POWER & ENERGY

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= F1 ⋅ dl ⋅ sin α

∵  

= F1 ⋅ ds ⋅ sin α π θ  = F1 sin  +  ⋅ r ⋅ dθ  4 2 π θ  = 16 sin  +  ⋅ dθ  4 2

63

! for infinitesimally small displacement dl ,  !  dl = ds(distance moved along path) 

 we have,  ⇒  ⇒ 

θ = s/r  s = r ⋅θ   ds = r ⋅ dθ 

! In the same time interval ‘dt’, work done by F2 is given as ! ! dw2 = F2 ⋅ dl

∵

= F2 ⋅ dl

! ! F2 & dl 

= F2 ⋅ ds ∴

= (20 − s) ⋅ ds ! Net work done by F1 is

w1 = ∫ dw1 π /2

= 16

π

θ

∫ sin  4 + 2  ⋅ dθ 0

π /2

−16 π θ  = cos  +  (1/2)  4 20

= −32 {cos(π /2) − cos(π /4)} = 16 2 J = 22.63 J

! and net work done by F2 is w2 = ∫ dw2 =

π r/2

(20 − s) ⋅ ds

0

= 20

π r/2 s0

s2 − 2

π r/2 0

π r π 2r 2 = 20 × − 2 8

= 105.8 J WORK, POWER & ENERGY

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! ! Work done by F3 : As F3 is a constant force, work done by F3 , w3 = F3 × displacement along F3 = F3 × r = 25 × 4 J = 100 J ∴

Change in K.E. of particle, ∆k = wall

k f − ki = w1 + w2 + w3

1 2 1 2 mvB − mv A = ( 22.63 + 105 ⋅ 8 + 100 ) J 2 2

vB =

2 × 228 ⋅ 43 + 100 5

= 13 ⋅ 83 m/s.

Example – 25 A man of height h0 = 2 m is bungee jumping from a platform situated a height h = 25 m above a lake. One end of an elastic rope is attached to his foot and the other end is fixed to the platform. He starts falling from rest in a vertical position. The length and elastic properties of the rope are chosen so that his speed will have been reduced to zero at the instant when his head reaches the surface of the water. Ultimately the jumper is hanging from the rope, with his head 8 m above the water. (i) find the unstretched length of the rope. (ii) find the maximum speed and acceleration achieved during the jump. Solution: (i)

fig. 5.53

Let us denote the elastic constant (spring constant) of the rope by k and its unstretched length by l0. The maximum length of the rope is l1 = h − h0 = 23 m, whilst in equilibrium it is l2 = (23 − 8)m = 15 m. Initially, and at the jumper’s lowest position, the kinetic energy is zero. If we ignore the mass of the rope and assume that the jumper’s centre of mass is half-way up his body, we can use conservation of energy to write 1 mgh = k (l1 − l0 ) 2. 2 In addition, in equilibrium, mg = k (l2 − l0 ). Dividing the two equations by each other we obtain a quadratic equation for l0 ,

l02 + 2(h − l1)l0 + (l12 − 2hl2 ) = l02 + 4l0 − 221 = 0, which gives l0 = 13 m. WORK, POWER & ENERGY

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(ii) When the falling jumper attains his highest speed, his acceleration must be zero, and so this must occur at the same level as the final equilibrium position (l = l2 ). Again applying the law of conservation of energy, 1 2 1 mv + k (l2 − l0 ) 2 = mg (l2 + h0 ), 2 2 where the ratio m/k is the same as that obtained from the equilibrium condition, namely,

m l2 − l0 = . k g Substituting this into the energy equation, shows that the maximum speed of the jumper is v = 18 ms −1 ≈ 65 km h −1. It is easy to see that his maximum acceleration occurs at the lowest point of the jump. Since the largest extension of the rope (10 m) is five times that at the equilibrium position (2 m), the greatest tension in the rope is 5 mg. So the highest net force exerted on the jumper is 4mg, and his maximum acceleration is 4g.

Example – 26 There are two conservative fields of forces: ! ! (i) F = ayiˆ; (ii) F = axiˆ + byjˆ. Are these forces conservative? Solution: Let us find the work performed by each force over the path from a certain point 1( x1, y1) to a certain point 2( x2, y2 ) : (i)

! ! dw = F ⋅ dl = (ayiˆ) ⋅ ( dxiˆ + dyjˆ ) = ay ⋅ dx x2

w = ∫ dw = a ∫ y ⋅ dx ; x1

! ! dw = F ⋅ dl = ( axiˆ + byjˆ ) ⋅ ( dxiˆ + dyjˆ )

(ii)

= ax ⋅ dx + dy ⋅ dy

x2

y2

x1

y1

w = ∫ dw = a ∫ x ⋅ dx + b ∫ y.dy

In the first case the integral depends on the type of y(x) function, that is, on the shape of the path. Consequently, the first field of force is not a conservative field. In the second case both integrals do not depend on the shape of the path: they are defined only by the coordinates of the initial and final points of the path: the second field of force is a conservative field.

WORK, POWER & ENERGY

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Example – 27 ! In a certain conservative field a particle experiences the force F = a( yiˆ + xjˆ) , where a is a constant, and iˆ and ˆj are unit vectors of the x and y axes respectively. Find the potential energy U ( x, y) of the particle in this field. ! Solution: Let us calculate the work performed by the force F over the distance from the point O (Fig.) to an arbitrary point P( x, y). Taking advantage of this work being independent of the shape of the path, we choose one passing through the points OMP and consisting of two rectilinear sections, then M

wOP =

! ! Fdr +

0

P

! !

∫ Fdr .

y

p(x, y)

j

M

The first integral is equal to zero since at all points of ! ! the OM section y = 0 and F ⊥ dr . Along the section ! ! ! MPx is constant, therefore, F ⋅ dr = F ⋅ ˆjdy

= F

y

0

i

M(x, 0) fig. 5.54

d y = a x d y and therefore, P

wOP = 0 + ax ∫ dy = axy. M

We know that this work must be equal to the decreases in the potential energy, i.e., wOP = U O − U P. Assuming U O = 0, we obtain U P − wOP , or U ( x, y) = −axy.

Another way of finding U ( x , y ) is to resort to the total differential of that function:

 ∂U dU =   ∂x

  ∂U  dx +    ∂y

  dy. 

Taking into account that ∂U/∂x = −Fx = −ay and ∂U/∂y = −Fy = −ax, we get dU = −a( ydx + xdy) = d (−axy).

hence

WORK, POWER & ENERGY

U(x, y) = –axy.

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EXERCISE

OBJECTIVE [ LEVEL - I ] Q. 1

Q. 2

Q. 3

Which of the following can be negative (a) kinetic energy (c) mechanical energy

(b) potential energy (d) energy.

Work done by force of friction (a) can be zero (c) can be positive

(b) can be negative (d) any of the above.

If force is always perpendicular to motion: (a) speed is constant (c) work done is zero

(b) velocity is constant (d) K.E. remains constant.

Q. 4

A block of mass m slides down a smooth vertical circular track. During the motion, the block is in (a) vertical equilibrium (b) horizontal equilibrium (c) radial equilibrium (d) none of these.

Q. 5

Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is 1 2 1 2 (a) kx (b) – kx 2 2 1 2 1 2 (c) kx (d) – kx . 4 4

Q. 6

A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be (wrt to the ground) (a) zero (b) mgvt cos²θ. (c) mgvt sin θ cos θ (d) mgvt sin2θ.

Q. 7

An elevator is moving upward with an acceleration a (and velocity v) when a man inside the elevator lifts a body of mass m through a height h (in time t). The average power developed by the man is (wrt elevator) m( g + a) h 1 (a) (b) m( g + a)(v + at ) t 2 mgh 1 (c) (d) mg (v + at ) . t 2 A particle is rotated in a vertical circle by connecting it to a string of length l and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is (a) gl (b) 2gl

Q. 8

(c)

3gl

WORK, POWER & ENERGY

(d)

5gl www.locuseducation.org

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Q. 9

68

A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is µ, the power delivered by the external agent after a time t from the beginning is equal to : (a) ma²t (b) µmgat (c) µ(a + µg)gt (d) m(a + µg)at.

Q. 10 The block of mass m is pulling, vertically up with constant speed, by applying force P. The free end of the string is pulled by l meter, the increase in potential energy of the block is : (a)

mgl 2

(b) mgl (c) 2 mgl (d)

P

mgl . 4

m

Q. 11 A block of mass M is pulled along a horizontal surface by applying a force at an angle θ with the horizontal. The friction coefficient between the block and the surface is µ. If the block travels at a uniform velocity, the work done by this applied force during a displacement d of the block is (a)

µMgd cos θ cos θ + µ sin θ

(b)

µMgd cos θ

(c)

µMgd sin θ cos θ + µ sin θ

(d)

µMgd cos θ sin θ

Q. 12 A spring placed horizontally on a rough horizontal surface is compressed against a block of mass m placed on the surface so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is µ, the potential energy stored in the spring is µ² m² g ² µ2 m2 g 2µm 2 g 2 3µ2 mg 2 (b) (c) (d) . (a) 2k 2k k k Q. 13 Work done in time t on a body of mass m which is accelerated uniformly from rest to a speed v in time t1 as a function of time t is given by 2

1 v 2 v 2 1 v2 2 1  mv  2 m t m t t (a) 2 t (b) (c)  (d) m 2 t .  t1 2 t1 2  t1  1 Q. 14 A block of mass m moving with a velocity v0 on a smooth horizontal floor collides with a light spring of stiffness k that is rigidly fixed horizontally with a vertical wall. If the maximum force imparted by the spring on the block is F, then: v0 k (a) F ∝ m m

(b)

F∝

(c)

F∝ v0

(d)

None of these.

k

WORK, POWER & ENERGY

B m x

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Q. 15 The kinetic energy acquired by a mass m in traveling a certain distance d, starting from rest, under the action of a constant force is directly proportional to (a)

(b) independent of m

m

(c) 1/ m

(d) m.

Q. 16 A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to (a) t 1/ 2 (b) t 3/ 4 (c) t 3/ 2

(d) t 2 .

Q. 17 Potential energy function U(r) corresponding to the central force F= K/r² would be (a) K r (c) Kr²

(b) K/r (d) K/r².

Q. 18 A motor drives a body along a straight line with a constant force. The power P developed by the motor must vary with time t as figure. Y

(a)

Y

Y

(b) P

P O

t

X

Y

(c) P

O

t

X

(d) P

O

t

X

.

O

t

X

Q. 19 A particle at rest on a frictionless table is acted on by a horizontal force which is constant in size and direction. A graph is plotted of the work done on the particle W, against the speed of the particle v. If there are no frictional forces acting on the particle, how graph will look like. Y

(a)

Y

(b) W

W O

Y

v

X

O

Y

(c) W v

X

(d) W

O

v

X

O

. v

X

! Q. 20 A particle is acted upon by a conservative force F = 7iˆ − 6 ˆj N (no other force is acting on the particle).

(

)

Under the influence of this force particle moves from (0, 0) to (–3m, 4m), then (a) (c)

work done by the force is 3 J (b) at (0, 0) speed of the particle must be zero (d)

WORK, POWER & ENERGY

work done by the force is –45 J at (0, 0) speed of the particle must not be zero

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OBJECTIVE [ LEVEL - II ] Q. 1

The potential energy of a particle varies with position x according to the relation U ( x ) = x 3 − 4 x. The point x = 2 is a point of (a) stable equilibrium (b) unstable equilibrium (c) neutral equilibrium (d) none of the above

Q. 2

The K.E. of a body moving along a straight line varies with time as shown in the figure. The force acting on the body is KE

(a) zero (b) constant (c) directly proportional to velocity (d) inversely proportional to velocity Q. 3

A ball is projected vertically upwards with an initial velocity. Which of the following graphs best represents the K.E. of the ball as a function of time from the instant of projection till it reaches the point of projection? Y

Y

Y

Y

(a) K.E.

(b) K.E.

(c) K.E.

( d) K.E.

O

Q. 4

Q. 5

Q. 6

t

t

X

O

t

X

O

X

t

O

. t

X

A small spherical ball is suspended through a string of length l. The whole arrangement is inside a vehicle which is moving with velocity v. Now suddenly the vehicle stops and ball starts moving along a circular path. If tension in the string at the highest point is twice the weight of the ball then (a) υ = 5gl

(b) υ = 7gl

(c) velocity of the ball at highest point is gl

(d) velocity of the ball at the highest point is 3gl

A particle with total energy E moves in one dimension in a region where the potential energy is U(x). The acceleration of the particle is zero where(a) U ( x ) = E

(b) U ( x ) = 0

(c) dU ( x ) = 0 dx

d 2U ( x ) =0 (d) dx 2

A block of mass 1 kg slides down a curved track that is one quadrant of a circle of radius 1 m. Its speed at the bottom is 2 m/s. The work done by the frictional force is (g = 10 m/s2) (a) (b) (c) (d)

–8 J +8J 9J –9J.

WORK, POWER & ENERGY

R =1m

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Q. 7

Q. 8

71

A mass of M kg is suspended by a weightless string. The minimum horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is : (a)

Mg ( 2 − 1)

(b)

Mg ( 2 + 1)

(c)

mg 2

(d)

Mg . 2

A ring of mass m can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant k = 4 mg/R, where 2R is the natural length of the spring. The other end of spring is fixed to the ground at a horizontal distance 2R from the base of the rod. If the mass if released at a height 1.5 R, then the velocity of the ring as it reaches the ground is : m

Q. 9

(a)

gR

(b)

2 gR

(c)

2gR

(d)

3gR .

1.5 R

2.0 R

A particle which is constrained to move along the X-axis is subjected to a force in the same direction which varies with the distance x of the particle from the origin as f(x) = – kx + ax³. Here k and a are positive constants. For x > 0, the functional form of the potential energy U(x) of the particle is [JEE] U(x)

U(x)

x

(a)

(b)

U(x)

x

U(x )

x

(c)

(d)

x

.

! Q. 10 The potential energy function associated with the force F = 4 xyiˆ + 2 x 2 ˆj is :

(a)

U = −2 x 2 y

(b)

U = −2 x 2 y + constant

(c)

U = 2 x 2 y + constant

(d)

not defined

Q. 11 An automobile engine of mass M accelerates and a constant power P is applied by the engine. The distance x covered in time t is given by 1/ 2

(a)

 8 Pt 3  x=   9M 

(c)

 Pt 3  x=   9M 

1/ 2

(b)

 8 Pt 3  x=   M 

(d)

 Pt 3  x=  M 

1/ 2

WORK, POWER & ENERGY

1/ 2

.

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Q. 12 In the system shown in the figure the mass m moves in a circular arc of angular amplitude 60° and radius r. 4m is stationary. Then 4m 60°

m A

m B

(a) (b) (c) (d)

the minimum value of coefficient of friction between the mass 4 m and the surface of the table is 0.50 the work done by gravitational force on the block m is positive when it moves from A to B the power delivered by the tension when m moves from A to B is zero. the kinetic energy of m in position B equals the work done by gravitational force on the block when it moves from position A to position B. Q. 13 A particle of mass m moves on the x-axis under the influence of a force of attraction towards the origin O given by F = –k/x² iˆ . If the particle starts from rest at a distance a from the origin the speed it will attain to reach the origin will be : (a)

2k  a − x  m  ax 

(c)

k  ax  m  a − x 

(b)

2k  a + x  m  ax 

(d)

m a − x . 2k  ax 

12

12

12

12

! Q. 14 The potential energy for a force field F is given by U ( x, y ) = cos ( x + y ) . The force acting on a particle  π at a position given by coordinates  0,  is  4

(

)

1 ˆ ˆ i+j 2 1 3 ˆ ˆ (c)  2 i + 2 j   

(a) −

(

1 ˆ ˆ i+ j 2 1 3 ˆ (d)  2 i − 2  (b)

)  ˆj   

Q. 15 A machine delivers constant power to a body moving along a straight line. The distance moved in time ‘t’ increases by what factor from to t = 2 to t = 8 sec. (a) 8

(b) 4

(c) 4 2

(c) 2 2

Q. 16 A partical moves in a circular path of radius r and its centripetal acceleration is given by : ac = k 2 rt 2 here k is a constant. The power delivered to the particle is (a) 2π m k 2 r 2

(b) m k 2 r 2t

m k 4 r 2t 5 (c) 3

(d) 0

WORK, POWER & ENERGY

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Q. 17 If wall = ∆kE then where does the ∆PE (if any) go? (a) Not accounted in the formulae (b) We should add an extra term (c) Has been taken care of in RHS (d) Has been taken care of in LHS

v

Q. 18 A stone is thrown off a cliff at a speed v at an angle θ with the horizontal. The plot of its speed with which if hits the earch as a function of θ is: v

(a)

–90º

θ

v θ +90º 180º 270º

(b)

+90º

v

180º

270º

θ

v

(c)

(d) θ

θ

Q. 19 A ball is tied to a string and given a horizontal velocity as shown. When will the string slack? (a)

(c)

 2 gl − u 2  θ = cos    3gl 

(b)

 2 u2  θ = cos  −   3 6 gl 

(d)

−1

−1

 2 gl − u 3  θ = cos −1    4 gl 

θ

 4 u2  θ = cos −1  −   3 3gl 

u

Q. 20 A block (B) is attached to two unstretched springs S1 and S2 with spring constant k and 4k, respectively. The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and support have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance ‘x’ and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. Find 2

M1

M2 S2

2

M2

y . x

1

S1

B

x S2

S1

M1

1

B x

(a)

4

WORK, POWER & ENERGY

(b) 2

(c)

1 2

(d)

1 4

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SUBJECTIVE [ LEVEL - I ]

Q. 1

Non-deformable spheres are kept in a closed packed arrangement as shown. Another identical layer is kept on top such that the centre of each sphere in this layer is vertically above those in the bottom layer. The radius of each sphere is R and mass is m. Is this the most stable configuration in terms of gravitational potential energy?

Q. 2

A partical makes one complete loop in time ‘t’ and then comes to rest. You view it from a space ship that is accelerating downwards with acceleration ‘a’. What is the change in kinetic energy? Does this violate the work-energy theorem’s concept that work done by all forces is equal to the increment in kinetic energy.

Q. 3

m

a

The string is pulled slowly as shown till the block of mass ‘m’ makes as angle of 60º with the vertical. ! What is the work done by F ? Pulley shown in the figure is smooth. sm o o

F R

th m Initial Position

Q. 4

A body of 4kg mass is placed on a horizontal surface and experiences a force varying with distance as shown in the graph. Find the speed of the body when the force ceases to act on it. F(t)

30 20 10 O

Q. 5 Q. 6

Q. 7

x(mts) 1

2

3

4

5

6

7

8

Find the average frictional force needed to stop a car weighing 500 kg in a distance of 25 m if the initial speed is 72 km/h. ! ! An object is displaced from position vector r1 = (2iˆ + 3 ˆj)m to r2 = (4iˆ + 6 ˆj )m under a force ! F = (3x 2iˆ + 2 yjˆ) N . Find the work done by this force.

 x3  = U  − 4 x + 6  . Here, U is in joule If potential energy of a particle moving along x-axis is given by:  3  and x in metre. Find positions of stable and unstable equilibrium.

WORK, POWER & ENERGY

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Q. 8

75

Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring.

m k

Q. 9

If the system is released from rest, determine the speeds of both masses after B has moved 1 m. Neglect friction and the masses of pulleys A

36 kg

Stationary Pulley

B

4kg

Q. 10 A block of mass m attached with an ideal spring of force constant k is placed on a rough inclined plane having inclination θ with the horizontal and coefficient of friction 1 µ = tan θ . Initially the block is held stationary with the 2 spring in its relaxed state, find the maximum extension in

k m θ

the spring if the block is released. Q. 11 The system shown in figure is in equilibrium. Determine the acceleration of all the loads immediately after the lower thread keeping the system in equilibrium has been cut. Assume that the threads are weightless, the mass of the pulley is negligibly small and there is no friction at the point of suspension. Also find the elastic potential energy stored at this moment in the left string. Q. 12 A smooth table is placed horizontally and a spring of unstretched length l0 and force constant k has one end fixed to its center. To the other end of the spring is attached a mass m which is making n revolutions per second around the center . Show that the radius r of this uniform circular motion is kl0 /( k − 4π 2 mn 2 ) and the tension T in the spring is 4 π 2 mkl0 n 2 /( k − 4π 2 mn 2 ) .

WORK, POWER & ENERGY

m1

m3 k

m2

k m4

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Q. 13 A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5.0 cm (figure). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring ? Q. 14 A block A of mass m is held at rest on a smooth horizontal floor. A light frictionless, small pulley is fixed at a height of 6 m from the floor. A light in extensible string of length 16 m , connected with A passes over the pulley and another identical block B is hung from the string. Initial height of B is 5 m from the floor as shown in figure. When the system is released from rest, B starts to move vertically downwards and A slides on the floor towards right. (a) If at an instant string makes an angle θ with horizontal, find relation between speed u of A and v of B. (b) Calculate v when B strikes the floor.

P B A

θ

Q. 15 A object is attached to the lower end of a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amount d. If the same object is attached to the same vertical spring but permitted to fall instead, through what distance does it stretch the spring? Q. 16 The string in figure has a length l = 4.0 ft. When the ball is released, it will swing down the dotted arc (a circle with center at O). How fast will it be going when it reaches the lowest point in its swing? l

O

d

nail

Q. 17 The nail in figure is located a distance d below the point of suspension. Show that d must be at least 0.6l if the ball is to swing completely around in a circle centered on the nail. Q. 18 Consider the situation shown in figure. Mass of block A is m and that of block B is 2m. The force constant of spring is K. Friction is absent everywhere. System is released from rest with the spring unstretched. Find: (a)

the maximum extension of the spring xm

(b)

the speed of block A when the extension in the spring is x =

(c)

A

B

xm 2 x net acceleration of block B when extension in the spring is x = m . 4

WORK, POWER & ENERGY

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Q. 19 In the figure the block A is released from rest when the spring is in its natural length. For the block B of mass m to leave contact with the ground at some stage what should be the minimum mass of the block A?

Q. 20 A smooth narrow tube in the form of an arc AB of a circle at centre O and of radius r is fixed so that A is vertically above O and OB is horizontal. Particles P of mass m and Q of mass 2m with a light inextensible string of length (πr/2) connecting them are placed inside the tube with P at A and Q at B and released from rest. Assuming the string remains taut during motion, find the speed of particles when P reaches B.

WORK, POWER & ENERGY

A

B m

A P r

O

Q

B

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[ LEVEL - II ] Q. 1

Q. 2

R

There is a fixed frictionless ring and three strings pass over it symmetrically as shown. The system is released from rest when the three strings are in the horizontal plane. What is the work done by gravity after sufficient time has passed? It is given that M > 3m. Assume there is no friction anywhere and all collisions cause the colliding bodies to stick together.

120°

m

m

M

1m

The spring is initially horizontal and its natural length is 0.317 mts. It is free to slide on a cylinder fixed perpendicular to the wall, as shown. The joint at A is a spherical joint capable of rotating in all directions. The block is released and allowed to come to rest. What is the work done on the block by the wall w? (g = 1 m/s2) The spring constant is 20 N/m. The wall has coefficient of friction, 0.1.

Q. 3

l

m

1m

1m

frictionless cylinder A

1m

5kg Cylinder is fixed perpendicular to the surface of the wall

W

You whirl a stone attached to a string in a horizontal circle as shown. What is the net work done by tension on the stone when the stone is brough to this state from the state of rest in vertical position? g = 1 m/s2, L = 2 mts, m = 3 kg. ω = 1 rad/s2

L

m

Q. 4

In the system, the natural length of each spring is 1 mt and they are both released from rest at the instant shown. Plot the velocity of the block at point A as a function of a ‘m’. What is its value as m → 0 ? Is this what you would expect to observe partically? If not, why?

WORK, POWER & ENERGY

A 1m k=10N/m2 m 2 mts www.locuseducation.org

PHYSICS

Q. 5

Q. 6

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In the figure shown, dimensions of the upper block are negligible compared to the lower one which has length L ‘L’ and is moving as with velocity ‘v’ on a frictionless surface. Plot the energy wasted by friction as a function m of ‘µ’ the friction coefficient between the blocks, if the upper block is kept on it jently with zerospeed with v M µ respect to the ground. ( M >> m ). An object is spun in a vertical circle as shown. What is the minimum speed ‘u’ required for it to enter region - III?

III

II

IV

I u

L 8

Q. 7

Q. 8

Q. 9

A ball of mass ‘m’ is released from angle ‘θ ’ as shown and meets a peg ‘P’ that is fixed on the wall as shown and moves in a circle there after. What is the minimum value of θ for it to complete the circle if the dimensions of the peg are negligible?

θ

L

L 2

P M

L 2 R m

A sphere of mass m held at a height 2R between a wedge of same mass and a rigid wall, is released. Assuming that all the surfaces are frictionless, find the speed of the sphere when the sphere hits the ground.

2R

m α

A ring of mass m = 1.2 kg can slide over a smooth vertical rod. A light string attached to the ring is passing over a smooth fixed pulley at a distance of L = 0.5 m from the rod as shown in the figure. At the other end of the string mass M = 3 kg is attached, lying over a smooth fixed inclined plane of inclination angle 37°. The ring is held in level with the pulley and released. Determine the velocity of the ring when the string makes an angle (α = 37°) with horizontal. L

M

m

37°

37°

WORK, POWER & ENERGY

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Q. 10 Two bars of masses m1 and m2 connected by a non-deformed light spring rest on a horizontal plane. the coefficient of friction between the bars and the surface is equal to µ. What minimum constant force has to be applied in he horizontal direction to the bar of mass m1 in order to shift the other bar? Q. 11 The potential energy corresponding to a certain two-dimensional force field is given by 1 U ( x, y) = k ( x 2 + y 2 ). (a) Derive Fx and Fy and describe the vector force at each point in terms of its 2

cartesian coordinates x and y. (b) Derive Fr and Fθ and describe the vector force at each point in terms of the polar coordinates r and θ of the point. (c) Can you think of a physical model of such a force? Q. 12 The system of mass A and B shown in the figure is released from rest with x = 0, determine the velocity of mass B when x = 3 m. 4m

4m

x

m √2

A m

B

4

2

x x Q. 13 The potential energy of a 2 kg particle free to move along the x-axis is given by U ( x) =   − 5   J b  b  where b = 1m. Plot this potential energy, identifying the extremum points. Identify the regions where particle may be found and its maximum speed. Given that the total mechanical energy is (i) 36; (ii) –4 J. Q. 14 A particle of mass m is kept on a fixed, smooth sphere of radius R at a position where the radius through the particle makes an angle of 30° with the vertical. The particle is released from this position. (a) What is the force exerted by the sphere on the particle just after the release ? (b) Find the distance travelled by the particle before it leaves contact with the sphere. Q. 15 A particle of mass m is kept on the top of a smooth sphere of radius R. It is given a sharp impulse which imparts it a horizontal speed v. (a) Find the normal force between the sphere and the particle just after the impulse. (b) What should be the minimum value of v for which the particle does not slip on the sphere ? (c) Assuming the velocity v to be half the minimum calculated in part (b), find the angle made by the radius through the particle with the vertical when it leaves the sphere. y Q. 16 A particle of mass ‘m’ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular ! path is ‘a’. A force F = yiˆ − xjˆ newton acts on the particle, where x, y denote the coordinates of position of the particle. Calculate the work done by this force in taking the particle from point A (a, 0) to point B(0, a) along the circular path. WORK, POWER & ENERGY

B(0, a) m

O

(x, y)

A(a, 0)

x

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Q. 17 Which of the following is/are conservative force (s)? Provide explanation/proof too. ! 5 ! (a) F = 2r 3rˆ (b) F = − rˆ r ˆ ˆ ! 3 xi + yj ! 3 yiˆ + xjˆ 3/ 2 3/ 2 (c) F = 2 (d) F = 2 ( x + y2 ) ( x + y2 )

(

(e)

)

(

)

! F = ayiˆ

Q. 18 A wedge of mass M with a smooth quarter circular plane, is kept on a rough horizontal surface. A particle of mass m is released from rest from the top of the wedge as shown in the figure. When the particle slides along the quarter circular plane, it exerts a force on the wedge. The wedge begins to slide when the particle exerts a maximum horizontal force on it. Find the coefficient of friction between the wedge & the horizontal surface. Q. 19 A triangle loop ABC of light string is passed over two small frictionless pulleys A and B. AB = BC = AC = 2t. Two masses m and M are attached to the mid point O of AB and point C respectively. If m is released, m & M cross each other at any point P as shown in the figure, then show that

m

M

O

A

m P

m 2 3− 5 5 > t. and OP = M 5 2

3 M each at 2 its top hangs from a ceiling by an inextensible string. If the rings gently pushed horizontally in opposite directions, find the angular distance covered by each ring when the tension in the string vanishes for once during their motion.

B

M

C

Q. 20 A loop of mass M with two identical rings of mass

WORK, POWER & ENERGY

m

m

M

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TR Y YOURSELF - I TRY ANSWER 1. 3. 5. 7. 9.

(a),(c),(d) 1  −  as 2 + bs  2  (d) zero

2.

1650 J

4.

3.33 × 10 −3 J

6. 8.

(a) –10J, (b) +10J zero

3 mv 2 4 x2

TR Y YOURSELF - II TRY ANSWER 1. 3. 5. 7.

(d) (b) 3.7 m/s (b)

2. 4. 6.

(a) 16 J mgl (1– cosθ )

9.

(a) –260 J

10.

1 ma 2 d 2

(b) (c) (d) (e)

300 J –40 J zero zero 1/2

11.

56 J

12.

13.

(a) (b) (c) (d) (e)

14.

yes; yes

–mgr (1– cos θ ) mgr (1 – cos θ ) 2g (1 – cosθ ) ; g sin θ cos–1 (2/3) at a greater angle

WORK, POWER & ENERGY

15.

2 µ mg +  4 µ 2 m2 g 2 − 4k ( µ 2 − 2 µ mgl0  2k

2 R [a sin θ + g (1 − cos θ ]

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83

TR Y YOURSELF - III TRY ANSWER 1.

mgl (1– cosθ )

2.

(a)

3.

(a), (d)

4.

(a)

5.

(a) V02 + 2 gl (1 − cos θ )

65mg ; (b) 3R

(b) 6. 7.

2 gl cos θ

mgR 2 l  sin   l R 2 R {a sin θ + g (1 − cos θ )}

8.

mg (sin θ + µ s cos θ )2 2k

9.

(a)

5gR

(b) P is above the horizontal by sin −1 (1/ 3) 10.

11.

mgR 2 l  sin   (a) l R (b)

l mgR 2 sin θ0 − sin (θ + θ0 ) + sin θ  where θ 0 = R l

(c)

gR l (1 − cos θ 0 ) where θ0 = R l

(a) N = mg (3cos θ − 2) 2 (b) For θ ≤ cos −1   : N B = 0, 3

N A = mg (3cos θ − 2 ) and 2 for θ ≥ cos −1   : N A = 0,  3

N B = mg ( 2 − 3cos θ ) . WORK, POWER & ENERGY

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TR Y YOURSELF - IV TRY ANSWER 2.

A/ r

3.

x=

2

b (stable) 2a Unstable

4.

Vertical Stable

5. 6.

7. 8. 9.

True (a) A –ve B –ve C +ve D +ve E zero (b) stable equilibrium : x = 6 unstable equilibrium : x = 2 (d) (c), (d) (a) 31.0 J (b) 5.33 m/s (c) Conservative

10.

(a) U ( x ) = − k

11.

ke2  1 1   −  (a) 2  R2 R1 

Horizontal

km1m2 d m1m2 if U (∞ ) = 0 (b) x ( d + x ) . 1 1 x

1 2 1 (b) −ke  −   R2 R1  1 2 1 1  (c) − ke  −  2  R2 R1  12.

(a) x ≥ a; (b) x ∈ φ ; (c) x ≤ a and x ≥ b; b a a b (d) − < x ≤ − and ≤ x < 2 2 2 2

WORK, POWER & ENERGY

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OBJECTIVE ANSWER [ LEVEL - I ] 1.

(b),(c),(d)

2.

(a),(b),(c),(d)

3.

(a),(c),(d)

4.

(d)

5.

(d)

6.

(d)

7.

(a)

8.

(c)

9.

(d)

10.

(a)

11.

(a)

12.

(a)

13.

(d)

14.

(a),(b),(c)

15.

(b)

16.

(c)

17.

(b)

18.

(a)

19.

(d)

20.

(b),(d)

OBJECTIVE ANSWER [ LEVEL - II ] 1.

(d)

2.

(d)

3.

(c)

4.

(b),(d)

5.

(c)

6.

(a)

7.

(a)

8.

(b)

9.

(d)

10.

(a),(b)

11.

(a)

12.

(a),(b),(c),(d)

13.

(a)

14.

(b)

15.

(a)

16.

(b)

17.

(d)

18.

(a)

19.

(a)

20.

(c)

WORK, POWER & ENERGY

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86

SUBJECTIVE ANSWER [ LEVEL - I ] 1.

NO

2.

3.

mgR 2

4.

5.

4000 N

6.

7.

Stable equilibrium at x = 2 and unstable equilibrium at x = –2.

8.

2mg k

9.

speed of B =

10.

mg sin θ k

11.

a1 = a2 = a3 = 0;

1 22 ma t 2 65 m /s

83 J

80 3 80 m /s , speed of A = m /s 85 2 85

 m + m2 − m3 − m4  a4 =  1  g; m4   (m2 g )2 Energy = 2k 13.

100 cm

14.

(a) v = u cos θ (b)

15.

2d

16.

16 ft/sec.

18.

(a)

19.

m 2

20.

4mg k

(b) 2 g

40 m /s 41

m 3k

(c)

g 2

2 (1 + π ) gr 3

WORK, POWER & ENERGY

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87

SUBJECTIVE ANSWER [ LEVEL - II ]

12.

1.

Mg R + 2 Rl − 3mgl

2.

10.03 Joules

3.

7.50 Joules

' 3.96 m/s.

2

U

13.

V

– 10

+ 10

4

x

25 4

4. (i) (ii)

m E mv 2 2

14.

(a)

5. v2 2gL

6.

7. 8. 9.

4

µ

 3 3 gL  2 +  2  

7   θ = sin −1  4 −  2 2 

2 gR .sin α

15.

m2    m1 +  µg 2  

11.

(a)

! Fx = −kx, Fy = −ky; F always points radially inward.

WORK, POWER & ENERGY

3mg 2

(b)

 −1  1  π  cos   −  ⋅ R ' 0.43R  3 6 

(a)

mg −

(b) (c)

gR cos−1 (7 / 12) ' 54.3°

mv 2 R

16.

Zero

17.

(a), (b), (c), (d).

18.

µ=

20.

cos −1 1

1.2 m/s

10.

–3 < x < 3; maximum speed = 6.5 m/s. E = – 4J is not possible

3m 3m + 2 M

( 3)

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